1 LUYN THI I HC CT 2012 CHNG DNG IN XOAY CHIUTP I Chng 3. Dng in xoay chiu : (9 cu).(1 cu): i cng v dng in xoay chiu: - Biu thc cng dng in xoay chiu v cc bi ton lin quan n thi gian.- T thng, sut in ng xoay chiu. - Cc i lng c trng cho dng in xoay chiu c s dng gi tr hiu dng, gi tr tc thi. (1 cu): Cc loi on mch in xoay chiu: - on mch in xoay chiu ch c in tr R. - on mch in xoay chiu ch c cun cm thun c h s t cm L- lch pha ca uRL v i.- Vit biu thc uRL, i, uL, uR.- Phng trnh lin h2 22 2 20 01.CCu iI Z I+ = ;2 22 2 20 01.LLu iI Z I+ = ;v cc h qu rt ra.- th ph thuc ca ZL theo L, ca uL theo i hoc ngc li.- on mch in xoay chiu ch c t in vi in dung C- lch pha ca uRC v i.- Vit biu thc uRC, i, uC, uR.- Phng trnh lin h 2 22 2 20 01.LCLCu iI Z I+ = v cc h qu rt ra.- th ph thuc ca ZC theo C, ca uC theo i hoc ngc li.(1 cu): Mch in xoay chiu RLC, hin tng cng hng in:- Vit biu thc u, i ca mch, in p gia cc phn t uR, uL, uC.- lch pha gia u v i, gia cc u thnh phn. - Hin tng cng hng in: cc c im v iu kin. - Mch in xoay chiu khi cun dy c thm in tr hot ng r 0. (1 cu): Cng sut ca mch in xoay chiu, h s cng sut:- Tnh cng sut ca mch in.- Tnh h s cng sut ca cc loi mch in. - Bi ton tnh gi tr ca cc i lng R, ZL, ZCkhi bit cng sut tiu th P. - Bi ton tnh cng sut, h s cng sut ca mch khi bitUR=mUL=nUChoc R=mZL=nZC. (2 cu): Cc tr trong mch in xoay chiu:- Mch in xoay chiu c R thay i- Mch in xoay chiu c L thay i- Mch in xoay chiu c C thay i- Mch in xoay chiu c (hoc f) thay i(1 cu): Bi ton bin lun hp kn, lch pha, gin vc t- Bi ton bin lun on mch c 1 hp kn.- Bi ton bin lun on mch c 2 hp kn.- Bi ton lch pha khi RL RCRLRCU UU UU U (1 cu): My bin p, s truyn ti in nng- My bin p: Tnh in p, s vng dy, cng dng in ca cun s cp v th cp.-Ch : Dng bi m cho c th l my tng p, hoc h p.- S truyn ti in nngTnh cng sut hao ph khi truyn ti.- Tnh gim in p.- Tnh hiu sut truyn ti in nng.(1 cu): Cc loi my pht in xoay chiu- My pht in xoay chiu 1 pha, 3 pha.- Cc s mc: hnh sao, hnh tam gic, biu thc lin h in p tng ng.- ng c khng ng b 3 pha. 2 o nBe DNG 1 I CNG V D XC A. Phng php : - Cho khung dy dn phng c N vng ,din tch S quay u vi vn tc e,xung quanh trc vung gc vi vi cc ng sc t ca mt t trng uc cm ng tB. 1. T thng gi qua khung dy : 0cos( ) cos( ) ( ) NBS t t Wb e e u = + = u +; T thng gi qua khung dy cc i 0NBS u=2. Sut in ng xoay chiu: - sut in ng cm ng xut hin trong khung dy: e=E0cos(et+0). tE0= NBeS - chu k v tn s lin h bi: 22 f 2 nTte= = t = tvi n l s vng quay trong 1 s - Sut in ng do cc my pht in xoay chiu to ra cng c biu thc tng t nh trn. - Khi trong khung dy c sut in ng th 2 u khung dy c in p xoay chiu . Nu khung cha ni vo ti tiu th th sut in ng hiu dng bng in p hiu dng 2 u on mch E = U 3.Khi nim v dng in xoay chiu- L dng in c cng bin thin tun hon vi thi gian theo quy lut ca hm s sin hay cosin, vi dng tng qut: i = I0cos(et + ) * i: gi tr ca cng dng in ti thi im t, c gi l gi tr tc thi ca i (cng tc thi). * I0 > 0: gi tr cc i ca i (cng cc i). * e > 0: tn s gc. f: tn s ca i. T: chu k ca i. * (et + ): pha ca i. * : pha ban u 4. Gi tr hiu dng: Ngoi ra, i vi dng in xoay chiu, cc i lng nh in p, sut in ng, cng in trng, cng l hm s sin hay cosin ca thi gian, vi cc i lng ny =02II

02UU=

02EE =5.Nhit lng to ra trn in tr R trong thi gian t nu c dng in xoay chiu i(t) = I0cos(et + i) chy qua l Q Q = RI2t Cng sut to nhit trn R khi c ddxc chy qua ; P=RI2 B.p dng : Bi1 : Mt khung dy dn phng c din tch S = 50 cm2, c N = 100 vng dy, quay u vi tc 50 vng/giy quanh mt trc vung gc vi cc ng sc ca mt t trng u c cm ng t B = 0,1 T. Chn gc thi gian t = 0 l lc vect php tuynn ca din tch S ca khung dy cng chiu vi vect cm ng t B v chiu dng l chiu quay ca khung dy. a)Vit biu thc xc nh t thnguqua khung dy. b)Vit biu thc xc nh sut in ng e xut hin trong khung dy. c)V th biu din s bin i ca e theo thi gian. Bigii : a)Khung dy dn quay u vi tc gc : = 50.2 = 100rad/s Ti thi im ban u t = 0, vect php tuynn ca din tch S ca khung dy c chiu trng vi chiu ca vect cm ng tB ca t trng. n thi im t, php tuynn ca khung dy quay c mt gc bng t e . Lc ny t thng qua khung dy l : ) cos( t NBS e | =Nh vy, t thng qua khung dy bin thin iu ho theo thi gian vi tn s gc v vi gi tr cc i (bin ) l 0 = NBS. Thay N = 100, B = 0,1 T, S = 50 cm2 = 50. 10-4 m2 v = 100 rad/s ta c biu thc ca t thng qua khung dy l : ) 100 cos( 05 , 0 t t | =(Wb) b)T thng qua khung dy bin thin iu ho theo thi gian, theo nh lut cm ng in t ca Faraday th trong khung dy xut hin mt sut in ng cm ng. Sut in ng cm ng xut hin trong khung dy c xc nh theo nh lut Lentz : |.|

\| = = = =2cos ) sin( ') (te e e e ||t NBS t NBSdtdet Nh vy, sut in ng cm ng xut hin trong khung dy bin i iu ho theo thi gian vi tn s gc v vi gi tr cc i (bin ) l E0 = NBS. Thay N = 100, B = 0,1 T, S = 50 cm2 = 50. 10-4 m2 v = 100 rad/s ta c biu thc xc nh sut in ng xut hin trong khung dy l : 3 |.|

\| =2100 cos 5tt t t e(V)hay|.|

\| ~2314 cos 7 , 15tt e(V) c)Sut in ng xut hin trong khung dy bin i iu ho theo thi gian vi chu kh T v tn s f ln lt l : 02 , 01002 2= = =ttetTs; 5002 , 01 1= = =TfHz th biu din s bin i ca sut in ng e theo thi gian t l ng hnh sin c chu k tun hon T = 0,02 s.Bng gi tr ca sut in ng e ti mt s thi im c bit nh : 0 s,005 , 04 =T s, 01 , 02 =T s,015 , 043=T s,02 , 0 = T s,025 , 045=T s v03 , 023=T s : t (s)00,0050,010,0150,020,0250,03 e (V)015,70-15,7015,70 th biu din s ph thuc ca e theo t nh hnh trn H1: Bi3 : Dng in xoay chiu chy qua mt on mch c cng bin i iu ho theo thi gian c m t bng th hnh di y. a)Xc nh bin , chu k v tn s ca dng in. b) th ct trc tung ( trc Oi) ti im c to bao nhiu ? Bigii : a)Bin chnh l gi tr cc i I0 ca cng dng in. Da vo th ta c bin ca dng in ny l : I0 = 4 A. Ti thi im 2,5.10-2 s, dng in c cng tc thi bng 4 A. Thi im k tip m dng in c cng tc thi bng 4 A l 2,25.10-2 s. Do chu k ca dng in ny l : T = 2,25.10-2 0,25.10-2 = 2.10-2 s;Tn s ca dng in ny l : 5010 . 21 12= = =TfHz b)Biu thc cng dng in xoay chiu ny c dng :) cos(0 it I i e+ =Tn s gc ca dng in ny l : t t t e 100 50 . 2 2 = = = frad/s Ti thi im t = 0,25.10-2 s, dng in c cng tc thi i = I0 = 4 A, nn suy ra : 0 0) 0 . 100 cos( I Ii= + t Hay 14cos =|.|

\|+it Suy ra :4t =i rad .Do biu thc cng ca dng in ny l :) (4100 cos 4 ) (4100 cos0A t A t I i|.|

\| =|.|

\| =ttttTi thi im t = 0 th dng in c cng tc thi l : 2 2242) (40 . 100 cos00= = =|.|

\| =IA I ittA 83 , 2 ~A.Vy th ct trc tung ti im c to (0 s,2 2A). Bi4 : Biu thc cng dng in xoay chiu chy qua mt on mch l) )( 100 cos(0A t I i t = , vi 0 0 I >v t tnh bng giy (s). Tnh t lc 0 s, xc nh thi im u tin m dng in c cng tc thi bng cng hiu dng ? Bigii : t (10-2 s) i (A) 0 + 4 - 4 0,250,751,251,752,252,753,25 t (s) e (V) 0 + 15,7 - 15,7 0,005 0,015 0,025 0,010,02 0,03 H.1 i, u t i (t) u (t) 0 4 Biu thc cng dng in) )( 100 cos(0A t I i t = c dng dao ng iu ho. Do , tnh t lc 0 s, tm thi im u tin dng in c cng tc thi bng cng hiu dng 20II i = =cng ging nh tnhthi gian t tnh t lc 0 s, V pha ban u ca dao ng bng 0, ngha l lc 0 s th I angc gi tr i = I0, nn thi im cn tm chnh bng thi gian ngn nht I bin thintim m i = I0 n v tr c 20II i = = . Ta s dng tnh cht hnh chiu ca mt cht im chuyn ng trn u ln mt ng thng nm trong mt phng qu o l mt dao ng iu ho vi cng chu k gii Biton ny. Thi gian ngn nht i = I0 n v tr c 20II i = = .(t P n D) chnh bng thi gian vt chuyn ng trn u vi cng chu k i t P n Q theo cung trn PQ. Tam gic ODQ vung ti D v c OQ = A, 2AOD =nn ta c : 22cos = =OQODo Suy ra : 4to =rad.Thi gian cht im chuyn ng trn u i t P n Q theo cung trnPQ l : e eteo414= = = tTrong biu thc ca dng in, th tn s gc = 100rad/s nn ta suy ra tnh t lc 0 s th thi im u tin m dng in c cng tc thi bng cng hiu dng l : 4001100 . 4 4= = =ttetts TRC NGHIM VN DNG I CNG DDXC Bi 1. Pht biu no sau y l khng ng? A. in p bin i iu ho theo thi gian gi l in p xoay chiu. B. Dng in c cng bin i iu ho theo thi gian gi l dng in xoay chiu. C. Sut in ng bin i iu ho theo thi gian gi l sut in ng xoay chiu. D. Cho dng in mt chiu v dng in xoay chiu ln lt i qua cng mt in tr th chng to ra nhit lng nh nhau. Bi 2. Trong cc i lng c trng cho dng in xoay chiu sau y, i lng no khng dng gi tr hiu dng? A. in p . B. Cng dng in. C. Sut in ng. D. Cng sut. Bi 3. t mt in p xoay chiu c gi tr hiu dng U v tn s f thay i vo hai u mt in tr thun R. Nhit lng to ra trn in tr A. T l vi f2B. T l vi U2C. T l vi f D. B v C ng Bi 4. Chn Bi ng. Cc gi tr hiu dng ca dng in xoay chiu: A. c xy dng da trn tc dng nhit ca dng in. B. c o bng ampe k nhit.C. bng gi tr trung bnh chia cho2 . D. bng gi tr cc i chia cho 2. Bi 5:Mt khung dy dt hnh trn tit din S v c N vng dy, hai u dy khp kn, quay xung quanh mt trc c nh ng phng vi cun dy t trong t trng uBc phng vung gc vi trc quay. Tc gc khung dy le . T thng qua cun dy lct > 0 l: A.u= BS.B. u= BSsine . C.u= NBScose t.D. u= NBS. Bi 6. Mt dng in xoay chiu c cng 2 2 cos(100 / 6) = + i t t t(A. . Chn Bi pht biu sai. A.Cng hiu dng bng 2 (A) .B.Chu k dng in l 0,02 (s). C.Tn sl 100t.D.Pha ban u ca dng in l t/6. Bi 7. Mt thit b in xoay chiu c cc in p nh mc ghi trn thit b l 100 V. Thit b chu c in p ti a l:A. 100 V B. 100 2 VC. 200VD. 50 2 V Bi 8 : Hy xc nh p n ng .Dng in xoay chiui = 10 cos100t t (A),qua in tr R = 5O.Nhit lng ta ra sau 7 pht l :O i + I0A2 P Q (C) D 5 -U0Ou U0u N M A .500J.B. 50J . C.105KJ. D.250 J Bi 9: biu thc cng dng in li = 4.cos(100 t t -t /4) (A). Ti thi im t = 0,04 s cng dng in c gi tr l A. i = 4 AB. i = 2 2AC. i =2AD. i = 2 A Bi 10: T thng qua mt vng dy dn l( )22.10cos 1004t Wbttt| |u = + |\ .. Biu thc ca sut in ng cm ng xut hin trong vng dy ny lA.2sin 100 ( )4e t Vtt| |= + |\ .B.2sin 100 ( )4e t Vtt| |= + |\ . C.2sin100 ( ) e t V t = D.2 sin100 ( ) e t V t t =DNG 2 GII TON XC BNG MI LIN QUAN GIA DDDH V CHUYN NG TRN U A. Phng php : 1.Ta dng mi lin h gia dao ng iu ho v chuyn ngtrn u tnh. Theo lng gic :0u = U cos(t + )c biu din bng vng trn tm O bn knh U0 , quay vi tc gce , +C 2 im M ,N chuyn ng trn u c hnh chiu ln Ou l u,nhngN c hnh chiu ln Ou c u ang tng (vn tc l dng) , cn M c hnh chiu ln Ou c u ang gim (vn tc l m ) + Ta xc nh xem vo thi im ta xt in p u c gi tr u v ang bini th no ( v d chiu m ) ta chn M ri tnh gc MOA = ;cn nu theo chiu dng ta chn N vtnh NOA = theo lng gic2. Dng in xoay chiu i = I0cos(2tft + i) * Mi giy i chiu 2f ln * Nu cho dng in qua b phn lm rung dy trong hin tng sngdng th dy rung vi tn s 2f 3. Cng thc tnh thi gian n hunh quang sng trong mt chu k Khi t in p u = U0cos(et + u) vo hai u bng n, bit n ch sngln khiu U1. Gi t A l khong thi gian n sng trong mt chu k

4teAA=Vi 1 0A = M OU ;10cosUU A = ,(0 < A < t/2) B.p dng : Bi1 : Biu thc cng dng in xoay chiu chy qua mt on mch l) )( 100 cos(0A t I i t = , vi I0 > 0 v t tnh bng giy (s). Tnh t lc 0 s, xc nh thi im u tin m dng in c cng tc thi bng cng hiu dng ? Bigii : Biu thc cng dng in) )( 100 cos(0A t I i t =ging v mt ton hc vi biu thc li ) cos( t A x e =ca cht im dao ng c iu ho. Do , tnh t lc 0 s, tm thi im u tin dng in c cng tc thi bng cng hiu dng 20II i = =cng ging nh tnh t lc 0 s, tm thi im u tin cht im dao ng c iu ho c li 2Ax = . V pha ban u ca dao ng bng 0, ngha l lc 0 s th cht im ang v tr gii hn x = A, nn thi im cn tm chnh bng thi gian ngn nht cht im i t v tr gii hnx = A n v tr c li 2Ax = . Ta s dng tnh cht hnh chiu ca mt cht im chuyn ng trn u ln mt ng thng nm trong mt phng qu o l mt dao ng iu ho vi cng chu k gii Biton ny. Thi gian ngn nht cht im dao ng iu ho chuyn ng t v tr c li x = A n v tr c li 2Ax =(t P n D) chnh bng thi gian cht im chuyn ng trn u vi cng chu k i t P n Q theo cung trn PQ. UuOM'2M2M'1M1-UU001-U1Sng SngTt Tt 6 Tam gic ODQ vung ti D v c OQ = A, 2AOD =nn ta c : 22cos = =OQODo Suy ra : 4to =rad Thi gian cht im chuyn ng trn u i t P n Q theo cung trnPQ l : e eteo414= = = tTrong biu thc ca dng in, th tn s gc = 100rad/s nn ta suy ra tnh t lc 0 s th thi im u tin m dng in c cng tc thi bng cng hiu dng l : 4001100 . 4 4= = =ttetts Bi2 (B5-17SBT NC)Mt n non mc vi mch in xoay chiu c in p hiu dng 220V v tn s 50Hz .Bit n sng khi in p gia 2 cc khng nh hn 155V . a) Trong mt giy , bao nhiu ln n sng ?bao nhiu ln n tt ? b) Tnh t s gia thi gian n sng v thi gian n tt trong mt chu k ca dng in ? Hng dn : a)220 2sin(100 )( ) u t V t = -Trong mt chu k c 2 khong thi gian tha mn iu kin nsng155 u >Do trong mt chu k ,n chp sng 2 ln ,2 lnn tt-S chu k trong mt giy : n = f = 50 chu k-Trong mt giy n chp sng 100 ln , n chp tt 100 ln b)Tm khong thi gian n sng trong na chu k u 220 2sin(100 ) 155 t t > 1sin(100 )2t t > 51006 6tt tt s s 1 5600 600s t s s s-Thi gian n sng trong na chu k : 5 1 1600 600 150t s A= = Thi gian n sng trong mt chu k :1 12.150 75t sS= =-Thi gian n tt trong chu k :1 1 150 75 150t T t stat s= = = -T s thi gian n sng v thi gian n tt trong mt chu k : 17521150tsttat= = C th gii Biton trnbng pp nu trn : 155 u > 220 21552= =02U. Vy thi gian n sng tng ng chuyn ng trn u quay gc EOM v gc' ' EOM . Biu din bng hnh ta thy tng thi gian n sng ng vi thi gian tS=4.t vi t l thi gian bn knh qutgc BOM =; vi 00/ 2 1cos2UU = = / 3 t = .p dng :4. / 3 14 / 300100 75St s stt= = = 1 / 7521 / 150tts ST tS ttat= = = O x + A A2 P Q (C) D CM M 02U 02U U0cosU0 OB

EE C 7 Bi3( H10-11): Ti thi im t, in p200 2 cos(100 )2u ttt = (trong u tnh bng V, t tnh bng s) c gi tr 100 2Vv ang gim. Sau thi im 1300 s , in p ny c gi tr l A. 100V. B. 100 3 . VC.100 2 . V D. 200 V. HD gii :Dng mi lin quan gia dddh v CDTD , khi t=0 , u ng vi CDTD C . Vo thi im t , u=100 2Vv ang gim nn ng vi CDTD ti M vi MOB = A .Ta c :100 2200 2uU A = =Suy ra teA= t=600.0,02/3600=1/300s . V vy thm 1300 s u ng vi CDTD B vi BOM =600. Suy ra lc u=100 2 . V Bi5: Vo cng mt thi im no , hai dng in xoay chiu i1 = Iocos(et + 1) v i2 = Iocos(et + 2) u cng c gi tr tc thi l 0,5Io, nhng mt dng in ang gim, cn mt dng in ang tng. Hai dng in ny lch pha nhau mt gc bng.A. 65tB. 32t C. 6tD. 34t Hng dn gii:Dng mi lin quan gia dddh v chuyn ng trn u : i vi dng i1 khi c gi tr tc thi 0,5I0 v ng tng ng vi chuyn ng trn u M , cn i vi dng i2 khi c gi tr tc thi 0,5I0 v ng gim ng vi chuyn ng trn u MBng cng thc lng gic chng dd c , ta c : '3MOB MOBt = = = 2'3MOMt= suy ra 2 cng dng in tc thi i1 v i2 lch pha nhau 23t P DNG : Bi1Dnginxoaychiuquamtonmchcbiuthc 0os(120 )3i I c t Att = .Thiimth2009 cng dng in tc thi bng cng hiu dng l: A. 120491440s B. 240971440s C. 241131440s D. p n khc Bi 2 ( 23 cc kho th ) in p tc thi gia hai u on mch240sin100 ( ) u t V t = . Thi im gn nht sau in p tc thi t gi tr 120V l : A.1/600sB.1/100sC.0,02s D.1/300s Bi 3: Dng in xoay chiu chy qua mt on mch c biu thc) 100 cos( 2 t t = t i A,ttnh bng giy (s).Dng in c cng tc thi bng khng ln th ba vo thi im A. ) (2005s . B.3( )100s .C. ) (2007s .D. ) (2009s . Cu4. Mt chic n nn t di mt in p xoay chiu 119V 50Hz. N ch sng ln khi in p tc thi gia hai u bng n ln hn 84V. Thi gian bng n sng trong mt chu k l bao nhiu? A. At = 0,0100s. B. At = 0,0133s. C. At = 0,0200s. D. At = 0,0233s. Bi 5 (H2007)Dng in chy qua mt on mch c biu thc i = I0cos100tt. Trong khong thi gian t 0 n 0,01s cng d tc thi c gi tr bng 0,5I0 vo nhng thi im A. 1400sv 2400s B. 1500sv 3500s C. 1300sv 2300sD. 1600sv 5600s. Bi6Dnginxoaychiuquamtonmchcbiuthc 0os(120 )3i I c t Att = .Thiimth2009 cng dng in tc thi bng cng hiu dng l: A. 120491440s B. 240971440sC. 241131440s D.pn khc. BC M U0 cos O B C C M 0,5I0 I0 cos O B C M 8 Bi 7 t in p xoay chiu c tr hiu dng U=120V tn s f=60Hz vo hai u mt bng n hunh quang. Bit n ch sng ln khi in p t vo n khng nh hn 60 2 V. Thi gian n sng trong mi giy l:A. 12s B. 13sC . 23sD. 14sBi 8inpgia haiu mtonmch cbiu thc 0os 1002u U c t Vtt| |= + |\ .. Nhng thiim t no sau y in p tc thi 02Uu = : A. 1400 s B. 7400 s C. 9400 s D. 11400 sBi 9 t in p xoay chiu c tr hiu dng U=120V tn s f=60Hz vo hai u mt bng n hunh quang. Bit n ch sng ln khi in p t vo n khng nh hn 60 2 V. T s thi gian n sng v n tt trong 30 pht l: A. 2 lnB. 0,5 lnC. 3 ln D. 1/3 ln Bi 10 Dng in chy qua mt on mch c biu thc i = I0cos100t. Trong mi na chu k, khi dng in chaichiuthkhongthigiancngdngintc thicgitr tuyti lnhnhocbng 0,5I0lA.1/300 s B. 2/300 sC. 1/600 s D 5/600s DNG 2 Biu thc in p xoay chiu. Biu thc cng dng in tc thi Tm gi tr tc thi ca i khi cho gi tr tc thi ca u v ngc li A. Phng php : - Vi mt on mch xoay chiu th biu thcin p hai u on mch v cng dng in qua mch c biu thc: u(t) = U0cos(et + u) i(t) = I0cos(et + i) Nu cho i =I0coset th 0u = U cos(t + ) Nu cho u =U0coset th0i = I cos(t - ) Nu cho u(t) = U0cos(et + u) i(t) = I0cos(et + u - )

- i lng = u - igi l lch pha gia u v i trong mt on mch. 0: > u sm pha hn i; 0: 0. Tnh t lc) ( 0 s t = , in lng chuyn qua tit din thng ca dy dn ca on mch trong thi gian bng na chu k ca dng in l A.0B.e02I C.et02ID.20etI HD:Ta c : 0, 5Tte=dqidt= 00. .cos( )2q i dt I ttete = = } }000sin( )22]I tIqtetee e= = .Bi4: Mt dng in xoay chiu c cng hiu dng l I c tn s lfth in lng qua tit din ca dy trong thi gian mt na chu k k t khi dng in bng khng l : A. 2 If t B. 2If t C. 2fIt D. 2fIt Bi5: Dng in xoay chiu hnh sin chy qua mt on mch c biu thc cng l) cos(0 it I i e+ = , I0 > 0. in lng chuyn qua tit din thng ca dy dn on mch trong thi gian bng chu k ca dng in l 12 A. 0.B. et02I.C. 20etI.D. e02I. Bi6: Dng in xoay chiu hnh sin chy qua mt on mch c biu thc c biu thc cng l |.|

\| =2cos0tet I i , I0 > 0. Tnh t lc) ( 0 s t = , in lng chuyn qua tit din thng ca dy dn ca on mch trong thi gian bng na chu k ca dng in l A. 0.B. et02I.C. 20etI.D. e02I. DNG 3 ON MCH CH C R ,L,C A. Phng php : 1. Mch in xoay chiu ch c tr thun u(t) = U0cos(et + ) ; u Ui = = 2cos(t + )R R.00U=Rv i , u cng pha.2. an mch ch c t in ; - T in cho dng in xoay chiu "i qua". - T in c tc dng cn tr dng in xoay chiu. - Gi s u =U0coseti = I0cos(et+ t/2) Cn i =U0cosetu = U0cos(et - t/2) Cn i =U0cos(et +i ) u = U0cos(et - t/2+i) -Dung khng:ZC -t ZC = 1C e; - Vy: nh lut m I = CUZ .**. ngha ca dung khng + ZC l i lng biu hin s cn tr dng in xoay chiu ca t in. + Dng in xoay chiu c tn s cao (cao tn) chuyn qua t in d dng hn dng in xoay chiu tn s thp. + ZC cng c tc dng lm cho i sm pha t/2 so vi u. 3.Mch in xoay chiu ch ccun cm :Mi cun dy c hai phn t : in tr r v t cm L . Ringcun cm thun ch c L - Trng hp nu rt li thp ra khi cun cm th sng n tng ln Cun cm c tc dngcn tr dng in xoay chiu. Tc dng cn tr ny ph thuc vo t cm cun dy. - Gi s i =I0coset u = LeI0cos(et+ /2) =U0cos(et+ t/2) Nu u =U0coset i =U0cos(et - t/2) i =I0cos(et+i) u = U0cos(et+ /2+i) - nh lut m: : I = LUe . -Cm khng ZL ZL = eL ngha ca cm khng + ZL l i lng biu hin s cn tr dng in xoay chiu ca cun cm. + Cun cm c L ln s cn tr nhiu i vi dng in xoay chiu, nht l dng in xoay chiu cao tn. + ZL cng c tc dng lm cho i tr pha t/2 so vi u. Lu : 1/10, 318 =t; 20, 636 =t ; 10,1592=t 2/ Cng thc tnh in dung ca t phng :C = dStc4 . 10 . 99 ( 11NC mi hc ) c : Hng s in mi.S: Phn th tch gia hai bn t (m3).d: Khong cch gia hai bn t(m). - in mi b nh thng l hin tng khi in trng tng vt qua mt gi tr gii hn no s llm cho in mi mt tnh cch in. - in p gii hn l in p ln nht m in mi khng b nh thng. B.p dng : Bi1 : t in p xoay chiu) )( 100 cos( 2 220 V t u t = , t tnh bng giy (s), vo hai u in tr thun R = 110 . Vit biu thc cng dng in chy qua in tr thun R. O RU Ix OIx CU 13 Bigii : Bin dng in xoay chiu chy qua in tr thun R l : 2 21102 22000= = =RUIA Dng in chy qua in tr thun R bin i iu ho cng tn s v cng pha vi in p xoay chiu gia hai u ca n nn biu thc ca dng in qua in tr thun R l : ) )( 100 cos( 2 2 A t i t = , t tnh bng giy (s) Bi2 : Biu thc cng dng in xoay chiu chy qua mt in tr thun R l ) (3100 cos 2 A t i|.|

\| =tt , t tnh bng giy (s). in p hiu dng gia hai du in tr thun o uc bng vn k xoay chiu l U = 150 V. a)Xc nh R. b)Vit biu thc in p gia hai u in tr thun R. Bigii : a)in tr thun R c xc nh t nh lut m : 1501150= = =IUR b)Bin ca in p xoay chiu gia hai u in tr thun R l : 2 150 20= =U UV in p gia hai u in tr thun R bin i iu ho cng tn s v cng pha vi dng in chy qua n nn biu thc in p gia hai u in tr thun R l : ) (3100 cos 2 150 V t u|.|

\| =tt , t tnh bng giy (s) Vn dng : Bi 1(C2007) Dng in xoay chiu trong on mch ch c in tr thun A. cng tn s vi in p hai u on mch v c pha ban u lun bng 0. B. c gi tr hiu dng t l thun vi in tr ca mch. C. cng tn s v cng pha vi in p hai u on mch. D. lun lch pha t/ 2 so vi in p hai u on mch Bi 2: Dng in xoay chiu trong on mch ch c in tr thun A. cng tn s vi in p hai u on mch v c pha ban u lun bng 0. B. cng tn s v cng pha vi in p hai u on mch. C. lun lch pha 2t so vi in p hai u on mchD. c gi tr hiu dng t l thun vi in tr ca mch. Bi2. Mt in tr thun R mc vo mt mch in xoay chiu tn s 50Hz. Mun dng in trong mch sm pha hn in pgia hai u on mch mt gc t/2. A. Ngi ta mc thm vo mchmt t in ni tip vi in tr. B. Ngi ta mc thm vo mchmt cun cm ni tip vi in tr. C. Ngi ta thay in tr ni trn bng mt t. D. Ngi ta thay in tr ni trn bng mt cun cm Bi3 : Mt on mch in xoay chiu ch c R=10, in p mc vo on mch l u =110 2 cos314t(V). Th biu thc ca cng dng in chy qua R c dng l: A.i =110 2 cos314t(A)B.i =110 2 cos(314t +2t)(A) C.i =11 2 cos314t(A)D.i =11cos314t(A) Bi4: t vo hai u in tr thun in p xoay chiu c gi tr hiu dng khng i, cho tn s dng in tng dn th cng dng in qua mch : A. Tng : B. Gim.C. Khng i . D. Tng n gi tr cc i sau gim. Bi 5 Mt on mch ni tip R,L,C c tn s dng in f = 50Hz; ZL=20O; ZC bin i c. Cho in dungC tng ln 5 ln so vi gi tr lc c cng hng in th gia in p u v cng i lch pha 3t. Gi tr ca R l: A.316 OB. 316O C.803 OD.316O 14 Bi 6( 22 cc kho th )Mch in xoay chiu R,C,Lni tip . t vo hai u on mch mt in p xoay chiu2 cos u U t e = .v lm thay i in dung ca t in th thy in p gia hai bn cc ca t in t gi tr cc i bng 2U .quan h gia cm khng ZL v in tr thun R l : A.ZL=R B.3LRZ = C. 3LZ R = D.ZL=3R Bi 7: tng dung khng ca t in phng c cht in mi l khng kh ta phi: A. gim in p hiu dng gia hai bn t in. B. tng tn s ca in pt vo hai bn t in. C. a thm bn in mi c hng s in mi ln vo trong lng t in. D. tng khong cch gia hai bn t in. Bi 8( H10-11): t in p u = U0coset vo hai u cun cm thun c t cm L th cng dng in qua cun cm l A. 0Ui cos( t )L 2t= e+e B. 0Ui cos( t )2 L 2t= e+e C. 0Ui cos( t )L 2t= eeD. 0Ui cos( t )2 L 2t= ee HD gii : V cng dng in qua on mch ch c L lun lun tr pha hn in p hai u mc mt gc 2 os( )( )2 2i I c t At te = DNG 4 TNG TRON MCH RLC NI TIP A. Phng php : Z =( )2Z Z2L CR+ gi l tng tr ca mch, V nh lut m : C MN R LL C MNU U U U UIZ R Z Z Z= = = = =vi M,N l hai im bt k. Hay cos = RZ. B.p dng : Bi1:Cho on mch gm in tr thun R = 100O v t in c in dung C mc ni tip. Bit biu thc in p gia 2 u on mch u = 100 cos 100tt Vv cng hiu dng trong mch I= 0,5 A. Tnh tng tr ca on mch v in dung ca t in? A. Z=100 2 O ; C= 1Zc e= F4101tB. . Z=200 2 O ; C= 1Zc e= F4101t C. Z=50 2 O ; C= 1Zc e= F4101tD. . Z=100 2 O ; C= 1Zc e=310Ft HD GII:Chn A. L m Z= U/I =100 2 O ;dng cng thc Z = 2 2 2 2100C CR Z Z + = +Suy ra ZC=2 2 2 22.100 100 100 Z R = = O ;C= 1Zc e= F4101t Bi2 Bitp 2tr 174 SGK: Mt on mch in AB gm mt in tr thun R = 100O, mt cun cm thun v mt t in mc ni tip (Hnh 33.2). Bit in p hiu dng gia hai u in tr, cun cm, t in ln lt l UR = 50 V,UL = 50 V, UC = 87,5 V, tn s dng in l 50 Hz. a) Tnh t cm ca cun cm v in dung ca t in. b) Tnh tng tr ca on mch AB v in p hiu dng UAB Bigii a) V I = R LU UR L=nn: Cm khng: ZL = LRURU = 100 O; t cm: L = LZ~e 0,318 H Tng t, ta c:- Dung khng: ZC = CRURU= 175 ;in dung ca t in: C = C1Z~e18,2 F. b) V AB l on mch RLC mc ni tip, nn c tng tr: Z = 2 2L CR (Z Z ) + = 125 OUAB = IZ = 50100.125 = 62,5 V 15 Bi3( H 2009): : t in p xoay chiu u = U0coset c U0 khng i vethay i c vo hai u on mch c R, L, C mc ni tip. Thay i e th cng dng in hiu dng trong mch khi e = e1 bng cng dng in hiu dng trong mch khi e = e2. H thc ng l : A. 1 22LCe+ e= .B. 1 21.LCee= .C. 1 22LCe+ e= .D. 1 21.LCee= . (HD:( ) ( )( )1 2 1 1 2 2 1 1 2 22 22 21 1 2 21 21 2 1 21 1 1 1 1. ....L C L C L C C LL C L CU UI I Z Z Z Z Z Z Z ZR Z Z R Z ZL LC Ce ee e ee= = = = + + | |+ = + = |\ . P DNG : Bi 7(C2007) Ln lt t in p xoay chiu u = 5 2 coset (V) vi e khng i vo hai u mi phn t: in tr thun R, cun dy thun cm (cm thun) c t cm L, t in c in dung C th dng in qua mi phn t trn u c gi tr hiu dng bng 50mA. t in p ny vo hai u on mch gm cc phn t trn mc ni tip th tng tr ca on mch l A. 300O B. 100OC. 100 2 O D. 100 3 O DNG 6Tnh in p hiu dng i vion mch RLC mc ni tip. : A. Phng php : X dng cng thc :( )22 2R L CU U U U = + hay RUcosU =hay .cosPUI = hay L CRU UtanU = hay dng gin vec tB.p dng : Bi1:Chn Bi ng. Cho mch in xoay chiu nh hnh v (Hnh 49). Ngi ta o c cc in p UAM = 16V, UMN = 20V, UNB = 8V. in p gia hai u on mch AB l: A. 44VB. 20VC. 28VD. 16V HD gii : Chn B. Dng cc cng thc: 2 2R L CU= U+(U-U); L CRU-Utg= U; Ucos= UR ; I =ZU; Io =ZUO.; UR = IR; UL = IZL; UC = IZC ; Bi2 ( Bi14-3 SBT)Cho mch in xoay chiu gmcun dy ( L,r) v C mc ni tip . in p 2 u on mch l240 2 cos100 ( ) u t V t =; R = 30O .T in c C thay i . Khi cho C c 2 gi tr C1 = 1310 Ft

v C2 =13.107Ft th cng nh nhau .Xc nh Udycaon mch cha cun dy ? HD: I1 = I2 Z1 = Z2 1 2402Z ZC CZL+= = O .Udy = 2 2. 50.4 2 200 2( ) R Z I VL+ = =Bi3( H 2009): : t mt in p xoay chiu c gi tr hiu dng U vo hai u on mch AB gm cun cm thun c t cm L, in tr thun R v t in c in dung C mc ni tip theo th t trn. Gi UL, UR v UC_ln lt l cc in p hiu dng gia hai u mi phn t. Bit in p gia hai u on mch AB lch pha 2tsoviinpgiahaiuonmchNB(onmchNBgmRvC).Hthcnodiyl ng? A. 2 2 2 2R C LU U U U = + + .B. 2 2 2 2C R LU U U U = + + . C. 2 2 2 2L R CU U U U = + + D. 2 2 2 2R C LU U U U = + +(HD : tng t Bi 8....lc ny UL max 2 2 2 2 2 2L NB R CU U U U U U = + = + + ) RLC AMNB Hnh 49 + L UR UC US P x O U16 Bi 5 (DH 2009):t mt in p xoay chiu c gi tr hiu dng U vo hai u on mch AB gm cun cm thun c t cm L, in tr thun R v t in c in dung C mc ni tip theo th t trn. Gi UL, UR v UC_ln lt l cc in p hiu dng gia hai u mi phn t. Bit in p gia hai u on mch AB lch pha 2tsoviinpgiahaiuonmchNB(onmchNBgmRvC).Hthcnodiyl ng? A. 2 2 2 2R C LU U U U = + + .B. 2 2 2 2C R LU U U U = + + . C. 2 2 2 2L R CU U U U = + + D. 2 2 2 2R C LU U U U = + +(HD : tng t Bi 8....lc ny UL max 2 2 2 2 2 2L NB R CU U U U U U = + = + + ) VN DNG : Bi 9: Cho mch in nh hnh v 1. in tr R=20O, cun cm L,tinC0.tgiaA,Bmtinpxoaychiunnh u=220 2 cos100t t(V)thtrongmchxyracnghngin vi gi tr hiu dng ca cng dng in l 5,5(A.. Xc nh in phiu dng gia Mv B. A. UMB=55V.B. UMB=110V.C. UMB=220V.D.UMB=440V. Bi 10.Cho mch in xoay chiu RLC nh hnh 4, cun dy thun cm. Bit UAF = 110(V), UEB = 112(V), UAB = 130(V). in phiu dng hai u t in c th nhn gi tr no sau y? A.88V. B.220V. C.200V. D.160V.Bi 11: Cho mch in nh hnh v. in tr R=20 O, cun cm L, t in C0. t gia A,B mt in pxoay chiu n nh u=220 2 cos100t t(V) th trong mch xy ra cng hng in vi gi tr hiu dng ca cng dng in l 5,5(A.. Xc nh in p hiu dng gia M v B. A. UMB=55V.B. UMB=110V.C. UMB=220V.D.UMB=440V. Bi 12.Cho mch in xoay chiu RLC nh hnh 4, cun dy thun cm. Bit UAF = 110(V), UEB = 112(V), UAB = 130(V). in phiu dng hai u t in c th nhn gi tr no sau y? A.88V. B.220V. C.200V. D.160V.Bi19(C2007)tinpu=U0costviU0vekhngivohaiuonmchRLCkhngphn nhnh. in p hiu dng hai u in tr thun l 80 V, hai u cun dy thun cm (cm thun) l 120 V v hai u t in l 60 V. in p hiu dng hai u on mch ny bng A. 220 V. B. 140 V.C. 100 V. D. 260 V. Bi 20: Cho mton mch xoay chiu ni tip gm in tr R, cun dy thun cm L v t C , t vo hai u on mch in p100 2 cos(100 ) u t V t = , lc 2L CZ Z =v in p hiu dng hai u in tr l UR = 60V . in p hiu dng hai u cun dy l: A. 60VB. 80VC. 120VD. 160V Bi 23 ( 21 cc kho th )Cho mch in nh hnh v : cun dy thun cm L ; vn k V1;V2 l vn k nhit c RV rt ln . t vo hai u A,B mt in p200sin( )( ) u t V e = + . Bit :1 2 C R e = ; L R e =. s ch ca vn k V1;V2 ln lt l : A.100 5 (V);100 5 (V)B.100 3 V;100V C.100 5 V;100VD.100 3 V;100 3 V DNG 7 LCH PHA gia in p v vi cng dng in xoay chiu A. Phng php : lch pha gia i v u :L C L CR1LU U Z ZCtanU R Re e = = =:ri suy rai lc ta x dngcos = RZ.hay RUcosU = ri suy ra , nh c th dng hay m+ Nu: ZL > ZC hay eL > C e1 th u nhanh pha hn i : >0(mch c tnh cm khng) A B C R L E F Hnh 4 A B C R L E F Hnh 4 B A MN V2 R L

V1C A R LC A B MN . ... Hnh 1 17 C AB R L NM C AB R L NM + Nu: ZL < ZC hay eL < C e1 th u chm pha hn i : ZL). vy uAN sm pha hn uMB 1 gc:132 4 2 4t t t t+ = + = . Da vo gin ta c:NB = 2MN v 14t=nn tam gic ANB l tam gic vung cn ti A v vy AB = AN Bi4( H 2009): : Mt on mch in xoay chiu gm in tr thun, cun cm thun v t in mc ni tip. Bit cm khng gp i dung khng. Dng vn k xoay chiu (in tr rt ln) o in p gia hai u t in v in p gia hai u in tr th s ch ca vn k l nh nhau. lch pha ca in p gia hai u on mch so vi cng dng in trong on mch lA. 4t.B. 6t.C. 3t.D. 3t . (HD: .22 2 tan ... 1R C L cR RU U Z Z RR = = = = = = =..) Bi5( H 2009): t in p u = U0coset vo hai u on mch mc ni tip gm in tr thun R, t in v cun cm thun c t cm L thay i c. Bit dung khng ca t in bngR 3 . iu chnh L in p hiu dng gia hai u cun cm t cc i, khi I R UC UL UAN UMB UA N B M 18 A. in p gia hai u in tr lch pha 6t so vi in p gia hai u on mch. B. in p gia hai u t in lch pha 6t so vi in p gia hai u on mch. C. trong mch c cng hng in. D. in p gia hai u cun cm lch pha 6t so vi in p gia hai u on mch. (HD: v gin vect ..chnh L ULmax dng nh l hm sin ta c : ( ) ( ) ( )max.sin ; sin ; sin ;LLRC RC L RC LU U UU H sU U U U U U= = = gc to bi ( );2RCU Ut= t ( );R RCU U o = vi tan 33cZRto o = = = in p gia hai u t in lch pha 2t so vi in p gia hai u on mch.&in p gia hai u cun cm lch pha 3t so vi in p gia hai u on mch. & in p gia hai u in tr lch pha 6t so vi in p gia hai u on mch nn A ng..) VN DNG : Bi 4( Bi2 Trang 174 NC )Cho mch in nh hnh v 174:Cho R = 100O; UR = 50V;UL = 50V ; UC = 87,5V. f = 50Hz( L , C ,fkhng i ) a) Tnh L ? C ?;b) Tnh Z ? UAB ? c) Tnh lch pha ca uAN v uMB ?HD : a) ZL = 100; ZC = 175 . b) Z = 125 ; UAB = 62,5 ( V ) c) 4ANt = ; 2MNt = ,34 2 4t t t A = + =Bi 5(C. 2010): t in p u=U0coset c e thay i c vo hai u on mch gm cun cm thun c t cm L, in tr thun R v t in c in dung C mc ni tip. Khi e < 1LC thA. in p hiu dung gia hai u in tr thun R bng in p hiu dng gia hai u on mch. B. in p hiu dng gia hai u in tr thun R nh hn in p hiu dng gia hai u on mch. C. cng dng in trong on mch tr pha so vi in p gia hai u on mch. D. cng dng in trong on mch cng pha vi in p gia hai u on mch. Bi 6(C. 2010): t in p xoay chiu vo hai u on mch gm in tr thun 40 O v t in mc ni tip.Bitinpgiahaiuonmchlchpha 3tsovicngdngintrongonmch.Dung khng ca t in bng A.40 3 OB. 40 33OC.40OD.20 3 O Bi 7(C. 2010): t in p 0u U cos( t ) (V)6t= e+vo hai u on mch gm in tr thun R v cun cm thun c t cm L mc ni tip th cng dng in qua on mchl 05i I sin( t ) (A)12t= e+ . T s in tr thun R v cm khng ca cun cm l A. 12.B. 1.C. 32.D.3 . Bi 8(C. 2010): t in p 0u U cos t = evo hai u on mch gm in tr thun R v t in C mc ni tip. Bit in p gia hai u in tr thun v in p gia hai bn t in c gi tr hiu dng bng nhau. Pht biu no sau y l sai ? A.Cng dng in qua mch tr pha4 tso vi in p gia hai u on mch. B.in p gia hai u in tr thun sm pha4 tso vi in p gia hai u on mch. C.Cng dng in qua mch sm pha4 tso vi in p gia hai u on mch. C AB R L NM Hnh v 174 19 Z Dung khng Cm . .khngZmin Of0f

D.in p gia hai u in tr thun tr pha4 tso vi in p gia hai u on mch. DNG 8 Cng hng in: A. Phng php : 1/- Nu U v R khng i th khi: ZL = ZC hay eL= C e1 th tngtr Z t gi trcc tiu Zmin = R, lc I t gi tr cc iI = Imax = U/R. Hin tng ny gil cng hng. 2/- ng cong cng hng ca onmch RLC.R cng ln thcng hng khng r nt3/- iu kin c cng hng l :1LCe= hay 2LC =1 hay2 2LC =1 4 f t hay1L CZ Z LCee= = Khi cng hngth:UR= U ; UL=UC ; UL,C=0 , 0 = uAB cngpha i ; uAB chm pha 2t so vi uL ; uAB nhanh pha 2t so vi uC 4/-. Lin h gia Z v tn s f : f0 l tn s lc cng hng . Khi f f0 : Mch c tnh cm khng , Z v f ng bin B.p dng : Bi1( H10-11): t in p u =2 cos U t evo hai u on mch AB gm hai on mch AN vNB mcnitip.on AN gm bintr R mc ni tip vi cun cm thunc t cm L, onNB ch c t in vi in dung C. t 112 LCe= . in p hiu dng gia hai u on mch AN khng ph thuc R th tn s gc e bng A. 1.2 2eB. 12. e C. 1.2eD. 2e1. HD gii : 2 22 22 2.( )LLANcR ZU UR Z Z+=+ 2 2 22 2 22 2 2 2( ) 2.. .( ) ( )L C L C cANL C L CR Z Z ZZ ZU U UR Z Z R Z Z+ = + + 22 2 22 22..( )L C cANL CZZ ZU U UR Z Z= + . UAN khng ph thuc R th : 2ZL=ZC Hay :2L.e=1/C.e haye=12LC=12. eHay :in p hiu dng hai u mch AN l : 2 22 2 22 2( ) 21AN AN LL C C L CLU UU IZ R ZR Z Z Z ZZR Z= = + =+ ++. UAN khng ph thuc vo R th Z2L-2ZCZL =0, suy ra 2 11 1(1).; (2)2 2 LC LCe e = = .Ly (1):(2). Ta c 2 12 e e =Bi 2 : Cho mch in xoay chiu AB gm R,L,C mc ni tip . Bit 2R C 16L = . on mch ang cng hng . bit in p hiu dng ca ton on mch AB l 120 V.Tnh in p hiu dng R L CU , U , U ? Hng dn : Gi thit cho :2R C 16L = 2R C 16 L e = e 2LC1R . 16ZZ= (1) v on mch cng hng nn :L CZ Z = (2) .t (1) v (2) 2 2 2L CR 16Z 16Z = = R L CU 4.U 4.U = = . Do U2=120VR L CU 120V; U U 30V = = =O 1LC e I R1 R2 > R1 20 Bi3( H10-11): t in p xoay chiu c gi tr hiu dng 200 V v tn s khng i vo hai u A v B ca on mch mc ni tip theo th t gm bin tr R, cun cm thun c t cm L v t in c in dung C thay i. Gi N l im ni gia cun cm thun v t in. Cc gi tr R, L, C hu hn v khc khng. Vi C = C1 th in p hiu dng gia hai u bin tr R c gi tr khng i v khc khng khi thay i gi tr R ca bin tr. Vi C = 12C th in p hiu dng gia A v N bng A. 200 V.B. 100 2 V.C. 100 V.D.200 2V. HD gii :Khi C = C1 th hiu in th hiu dng hai u bin tr l 2 2 21 12..( ) ( )1RL C L CU R UU I RR Z Z Z ZR= = =+ +. UR khng ph thuc vo R th ZL-ZC1=0( hin tng cng hng), suy raZC1 = ZL. Khi C=C1/2 , suy ra ZC=2ZC1=2ZL th in p hai u A v N l2 22 2 2 22 2 2 2. 200( ) ( 2 )LAN L LL C L LU R Z UU I R Z R Z U VR Z Z R Z Z+= + = + = = =+ + Hay :Theo gt Vi C = C1 th in p hiu dng gia hai u bin tr R c gi tr khng i v khc khng khi thay i gi tr R ca bin tr: mch cng hng Z L = Z C, C= C 12 Z C 1 = 2Z C U AN =U. (R2 +Z L2 ) / ((R2 +( Z L -Z C 1)2 ) =U . Chn ABi3 Cho on mch gm in tr R, t in C v cun t cm L mc ni tip (nh hnh v). Thay i tn s ca dng in xy ra hin tng cng hng in trong on mch th khng nh no sau y khng ng ? A. Cng hiu dng trong mch t cc i B. in p hiu dng gia cc im A, N v M, B bng nhau AN MBU U =C. in p hiu dng gia hai u on mch ln hn gia hai u in tr R D. Cng dng in tc thi qua mch ng pha in p tc thi gia hai u on mch VN DNG : Bi 1Hin tng cng hng trong mch LC xy ra cng r nt khi A. tn s ring ca mch cng ln. B. cun dy c t cm cng ln. C. in tr thun ca mch cng ln. D. in tr thun ca mch cng nh. Bi 2 Chn Bi sai trong cc Bi sau:Mch in xoay chiu RLC ni tip ang xy ra cng hng. Nu thay i tn s ca in p t vo hai u mch th: A. Cng hiu dng qua mch gim.B. H s cng sut ca mch gim. C. in p hiu dng trn R tng.D. Cng sut trung bnh trn mch gim. Bi 3Khi c cng hng in trong on mch xoay chiu RLC khng phn nhnh th : A. Cng dng in tc thi trong mch cng pha vi in p tc thi t vo hai u on mch B. in p tc thi gia hai u in tr thun cng pha vi in p tc thi gia hai bn t in C. Cng sut tiu th trn on mch t gi tr nh nht D. in p tc thi gia hai u in tr thun cng pha vi in p tc thi gia hai u cun cm Bi4:DungkhngcamtmchRLCmcnitipangcgitrnhhncmkhng.Munxyrahin tng cng hng in trong mch ta phi A. gim tn s dng in xoay chiu.B. tng in dung ca t in C. tng h s t cm ca cun dy.D. gim in tr ca mch. Bi 6 :Mch in gm R1,L1,C1 ni tip c tn s gc 1e v mch in gm R2,L2,C2 ni tip c tn s gc 2e . Bit 1e=2e v L1=2L2 .Hi on mch gm hai mch ni trn mc ni tip th cng hng khi tn se bng : A.1 21 2. eeee e +B.2 21 223e ee+= C.1 2. e e e =D.2 21 223e ee+=Bi 7 : t vo hai u mch RLC ni tip mt in p xoay chiu c gi tr hiu dng trn cc phn tR, L, C ln lt l 40V, 50V v 80V. Khi thay i tn s ca dng in mch c cng hng th in p hiu dng hai u in tr R bng A.50V.B.35V.C.70V.D.40V. L RC AMNB 21 CHNG DNG IN XOAY CHIU TP 2 DNG 9 Cng sut ca dng in xoay chiu.A. Phng php : 1 - Cng sut trung bnh : P = tW AA hay W = P.tvi AW l in nng tiu th trong khong thi gian At. - Suy ra cng sut tiu th ca dng in trong chu k Thay trong khong thi gian At >> T l: P= U.I.cosVi mch RLC ta c P = RI2

P = RI2 = ( )222L CU RR + Z - Z L,C, e =const, R thay i. R,C, e =const, Lthay i. R,L, e =const, C thay i. R,L,C,=const, f thay i. 2 2U UP= max22:RZ ZL CKhi R Z ZL C== Dng th nh sau: 2UP= max1:2RKhi Z Z LL CC e= = Dng th nh sau: 2UP= max1:2RKhi Z Z CL CL e= = Dng th nh sau: 2UP= max1:2RKhi Z Z fL CLC t= = Dng th nh sau: B.p dng : Bi 1 Cho mch in xoay chiu gm R,L,C ni tip .Bit in p 2 u mch :50 2cos100 ( ) u t V t =in p hiu dng UL = 30V ;UC = 60V. Bit cng sut tiu th trong mch P= 20W .Xc nh R,L,C ? HNG DN :UR = 2 2( ) C LU U U = 40V cos = URU = 0,8 . P= UIcos I = 0,5A; ZL = 60O ZC = 120O ; Bi 2: Chn cu ng. in pgia hai u mt on mch xoay chiu l:u=100 2 cos(100tt-t/6)(V) vcng dng inquamch li =4 2 cos(100tt-t/2)(A). Cng sut tiu th ca on mch l: A. 200W.B. 600W.C. 400W.D. 800W.HNG DN : Gii: CHN A. Dng. . os P U I c =.Vi =u -i=- t/6- (-t/2) = t/3 ; I= 4A; U =100V Bi 3:Cho mch in AB, trong C =F4104t, L=Ht 21, r = 25O mc ni tip.Biu thc in pgia hai u mchuAB= 50 2cos 100tt V .Tnh cng sut ca ton mch ? A. 50W B.25WC.100WD.50 2 W Hng dn gii : Chn ACng sut tiu th ca mch in : P = I2.r = 2.25=50 W, hoc : P =UICos Bi 6:Cho mch in xoay chiu RLC mc ni tip, c R l bin tr. t vo hai u on mch in pxoay chiu c biu thc120 2 cos(120 ) u t t = V. Bit rng ng vi hai gi tr ca bin tr :R1=18O,R2=32 O th cng sut tiu th P trn on mach nh nhau. Cng sut ca on mch c th nhn gi tr no sau y: A.144WB.288WC.576WD.282WHNG DN : Gii:CHN B . p dng cng thc:21 2( )L CR R Z Z = 1 224L CZ Z RR = = O + L UR UC US P x O UC O C0 P Pmax L O L0 P Pmax P R Pm OR1 Rm R2 P0 I=I1=I2 f1f2 f O f0 P 22 Vy 12 21 2 2 2 2 22288( ) ( )L C L CU UP R R WR Z Z R Z Z= = =+ +

Bi 7:Bi tp 1 tr 173sgk: on mch AB gm mt bin tr v mt t in c in dung C = 61,3 F mc ni tip. t vo hai u A, B mt in p xoay chiu u = 120cos100tt (V) Cn iu chnh cho in tr ca bin tr n gi tr no cng sut trn bin tr t cc i? Tnh gi tr cc i . Bi gii Trc ht, ta lp cng thc tnh cng sut tiu th in trn bin tr theo in tr R ca bin tr: 222 2CUP RI RR Z= =+ C th bin i: P = 22CUZRR+ .V R v 2CZR l cc s dng nn c th p dng bt ng thc C-si: (R + 2CZR) > 2ZC. ng thc xy ra khi R = ZC = 52 O. Thay vo cng thc tnh cng sut, ta tm c cng sut cc i trn bin tr: Pmax = 2CU69W2Z~Bi 8 (DH 2009) : t in p xoay chiu c gi tr hiu dng khng i vo hai u on mch gm bin tr R mc ni tip vi t in. Dung khng ca t in l 100 O. Khi iu chnh R th ti hai gi tr R1 v R2 cng sut tiu th ca on mch nh nhau. Bit in p hiu dng gia hai u t in khi R=R1 bng hai ln in p hiu dng gia hai u t in khi R = R2. Cc gi tr R1 v R2 l: A. R1 = 50O, R2 = 100 O.B. R1 = 40O, R2 = 250 O. C. R1 = 50O, R2 = 200 O.D. R1 = 25O, R2 = 100 O. (HNG DN ::2 2 1 21 2 1 1 2 2 1 2 1 2 2 2 2 21 2(1)... (2) & 2 2 (3)C CC CR RP P R I R I U U I IR Z R Z= = = = =+ + t (1) v (3) 2 14 (4) R R = th (4) vo (2) ta c :21 250 2004CZR R = = O = O) Bi 9: Mt on mch in gm in tr R ni tip vi cun dy thun cm c t cm L = t 41H. in p2 u on mch c gi tr hiu dng l 100V v c tn s f = 60Hz. Cngsut tiu th ca mch l 100W. Tnh RA.R= 10O hoc 90OB.R= 20O hoc 80OC.R= 90O D.R= 10O HNG DN : GII:Chn A. P = UIcos = I2R = 22ZR U hay P= 22 2LURR Z +Vi R l n s .Gii PT bc 2Bi 11:Cho mch in xoay chiu RLC mc ni tip, c R l bin tr. t vo hai u on mch in pxoay chiu c biu thc120 2 cos(120 ) u t t = V. Bit rng ng vi hai gi tr ca bin tr :R1=18O,R2=32 O th cng sut tiu th P trn on mach nh nhau. Cng sut ca on mch c th nhn gi tr no sau y: A.144WB.288WC.576WD.282WHNG DN : Gii:CHN B . p dng cng thc:21 2( )L CR R Z Z = 1 224L CZ Z RR = = O Vy 12 21 2 2 2 2 22288( ) ( )L C L CU UP R R WR Z Z R Z Z= = =+ +

Cu 13( H 2011-2012) : Mt on mch AB gm hai on mch AM v MB mc ni tip. on mch AM gm in tr thun R1 mc ni tip vi t in c in dung C, on mch MB gm in tr thun R2 mc ni tip vi cun cm thun c t cm L. t in p xoay chiu c tn s v gi tr hiu dng khng i vo hai u on mch AB. Khi on mch AB tiu th cng sut bng 120 W v c h s cng sut bng 1. Nu ni tt hai u t in th in p hai u on mch AM v MB c cng gi tr hiu dng nhng lch pha nhau 3t, cng sut tiu th trn on mch AB trong tr.hp ny bng C AB RL NM R r,L C BA 23 A. 75 W.B. 160 W.C. 90 W.D. 180 W. HNG DN :: - Khi cha ni tt, ta c: Z1 = R1 + R2 v (ZL = ZC.P1 = 21 2UR + R (1).- Khi ni tt, ta c: R1 = 2 22 LR+ Z(2) v tanMB=3L 2Z = 3R (3)Cng sut P2 = 21 22 21 2 LU (R + R )(R + R )+ Z(4) T (1); (3) 212U = 120 (W)3RP = (5) T (2); (3); (4)222U4RP = (6)T (5); (6)2 1390( )4P P W = =Cu 14:(C 2011-12)t in p u = U0cos t e ( U0 vekhng i) vo hai u on mch xoay chiu ni tip gm in tr thun, cun cm thun v t in c in dung iu chnh c. Khi dung khng l 100 O th cng sut tiu th ca on mch t cc i l 100W. Khi dung khng l 200 O th in p hiu dng gia hai u t in l 100 2V. Gi tr ca in tr thun l: A. 100 OB. 150OC. 160OD. 120O HNG DN :: Cng sut tiu th ca on mch t cc i : khi mch cng hng ZL=ZC=1002; 100 10maxUP W U RRO = = =UC=I./CZ =10/ 2 2 2. .200 100 2 400 2 2.100 1002 2 2 / 2(100) ( )U RZ R R RCR R Z ZL C = = + = O+ + Bi 16 Cho mt on mch RLC ni tip. Bit 31 10L (H), C (F)4= =t t. t vo hai u on mch mtin pxoay chiu c biu thc:u 120 2cos100 t(V) = t , t tnh bng s;vi R thay i c.1. iu chnh R = 80 O a) Tnh tng tr ca on mch. B. Vit biu thc cng dng in trong mch. 2. Cn iu chnh cho in tr ca bin tr n gi tr no cng sut trn bin trt cc i. Pmax? 3. Cn iu chnh cho in tr ca bin tr n gi tr no cng sut trn bin trt gi tr 72W GII : 1.a) Tng tr ca on mch Z =221R LC| |+ e |e\ .;ZL = 1L 100 . 100( ) e = t = Ot ;ZC = 31 1 440( )C 100 . 10t= = Oe t;Z= ( )2280 100 40 100( ) + = OB. Biu thc ca i c dng: i = I0cos(100 t ) t(A) ;00U 120 2I 1, 2 2(A)Z 100= = =tan = 1L100 40 3CR 80 4e e= =; 0, 2 ~ t rad Vy: i = 1, 2 2 cos(100 t 0, 2 ) t t (A) ;t tnh bng s. 2. Cng sut tiu th trn on mch: P = RI2 = R.22 2L CUR (Z Z ) + = 22L CU(Z Z )RR+(1) V R; 2L C(Z Z )R l cc hng s dng nn theo bt ng thc Csi: 2L C(Z Z )RR+L C2 Z Z > ;ng thc xy ra khi:R =L CZ Z = 100 - 40 = 60O Cng sut tiu th cc i:P = 2 2L CU U120W2 Z Z 2R= = 24 3.T (1):R2 - 2UPR + (ZL ZC.2 = 0 R2 - 200R + 3600 = 0 =>Vy R1 = 180O, R2 = 20 O Bi 17: Cho on mch xoay chiu nh hnh v. Biu thc in phai u on mch khng i: u=2602cos(100tt)(v). Cc gi tr: L=2/t (H), C=104/t (F),r=10(O), R thay i c. Ban u iu chnh R gi trR = R0 = 40(O).a. Tnh : Cng sut tiu th ca on mch, vit biu thc i. Cho tan(1,176)=2,4. b. Cho R thay i. Tm R cng sut tiu th ca on mch cc i. Gi thit L=2/t (H), C=104/t (F), r=10(O), R thay i c. Ban u iu chnh R gi tr R=R0=40(O). u=2602cos(100tt)(v). Kt lun:a. Tnh P v vit i? (tg(1,176) = 2,4). b. Tm R cng sut tiu th ca on mch cc i. Bi gii:a. Tnh P vit i: - T biu thc u ta suy ra: U = 260(v), e = 100t (rad/s),ZL = eL = 200(O), ZC = 1/(eC. = 80(O ). - Tng tr: Z =( ) ( )2 2L CR Z -Z r + + = 130(O ).- Cng hiu dng: I = U/Z = 2(A);-Cng sut: P = I2(R+r) = 200(W).- lch pha gia u v i : tg = (ZLZC./(R+r)=2,4 = 1,176(rad) (= 57,40). - Biu thc cng qua mch: i = 22cos(100tt 1,176)(A). b. Tm R cng sut cc i: - Ta c: P = I2(R+r) =( ) ( )22 2CU (R r) 12R r ZLZ++ + = ( ) ( )22 2CUR r Z(R r)LZ + + + = ( )( )222CUZR r(R r)LZ + ++ = ( )26760014400R 10(R 10)+ ++ ;-P l hm s theo bin ph (R+10). - T bt ng thc csi ta suy ra: P t gi tr cc i (Pmax) khi:(R +10) = 14400(R 10) + R = 110(O). - Kho st P theo R:+ Khi R=0 th P=P0~43,9W,+ Khi R =Rm = 110(O ) th P= Pmax ~ 281,7(W)+ Khi R +P 0. - th P theo R: t v . Bi 18( H 2009): : t in p xoay chiu c gi tr hiu dng khng i vo hai u on mch gm bin tr R mc ni tip vi t in. Dung khng ca t in l 100 O. Khi iu chnh R th ti hai gi tr R1 v R2 cng sut tiu th ca on mch nh nhau. Bit in p hiu dng gia hai u t in khi R=R1 bng hai ln in p hiu dng gia hai u t in khi R = R2. Cc gi tr R1 v R2 l: A. R1 = 50O, R2 = 100 O.B. R1 = 40O, R2 = 250 O. C. R1 = 50O, R2 = 200 O.D. R1 = 25O, R2 = 100 O. (HNG DN ::2 2 1 21 2 1 1 2 2 1 2 1 2 2 2 2 21 2(1)... (2) & 2 2 (3)C CC CR RP P R I R I U U I IR Z R Z= = = = =+ + t (1) v (3) 2 14 (4) R R = th (4) vo (2) ta c :21 250 2004CZR R = = O = O) Bi 19 ( H10-11): t in p xoay chiu c gi tr hiu dng khng i, tn s 50Hz vo hai u on mch mc ni tip gm in tr thun R, cun cm thun c t cm L v t in c in dung C thay i c. iu chnh in dung C n gi tr 4104Ft hoc 4102Ft th cng sut tiu th trn on mch u c gi tr bng nhau. Gi tr ca L bng A. 1.2HtB. 2. HtC. 1.3HtD. 3. Ht Hng dn gii : C AB RL P R Pm OR1 Rm R2 P0 25 22 2( )L CUP RR Z Z=+ 2 21 2 2 2 2 21 2( ) ( )L C L CU UP R P RR Z Z R Z Z= = =+ + 1 22C CLZ ZZ+= ; th s tm ZLL=3. Ht Hay : Theo gi thit khi C =C1 hoc C = C2 th P1 = P2 nn ta c: 2 2 2 2 2 2 2 2 1 21 2 1 2 1 23( ) ( )2C CL C L C LZ ZI R I R Z Z R Z Z R Z Z Z L Ht+= = + = + = =Bi 21:Cho mch in xoay chiu nh hnh v. t vo hai u on mch mt in pc : U=100(v), tn s f=50Hz. Cc gi tr L = (0,2)/t (H),C=104/t (F). Bit uAN v uMB lch pha t/2. Tnh R v cng sut tiu th ca mch. Gi thit U= 100(v), f = 50Hz. Cc gi tr L = (0,2)/t (H), C=104/t (F). Bit uAN v uMB lch pha t/2. Kt lun TnhR? Bi gii: - Ta c: e = 2tf = 100t(Rad/s),- ZL = eL = 20(O ),ZC = 1/(eC. = 100(O). - V gin :uAN = uR + uL L R AN U U U + =,uMB = uR + uC C R MB U U U + =.- T gin tam gic OAB vung ti O : (UL + UC.2 = U2AN +U2MB

hay : U2L + U2C +2ULUC = U2R + U2L + U2R +U2C ULUC = U2RhayIZL IZC = I2R2 hay R=C LZ Z ~ 4,7(O ). - Tng tr:Z = ( )22L CR Z -Z + ~ 91,6(O ).- Cng hiu dng qua mch: I = U/Z~ 1,09(A). - Cng sut: P = I2R~ 53(W).Cu 3 (Cu 17 24 cc kho th) on mch MP gm MN v NP ghp ni tip .in p v cng dng in tc thi ln lt c biu thc120cos100 ( )MNu t V t = ;120 3sin100 ( )NPu t V t = ; 2sin(100 )( )3i t Vtt = +. tng tr v cng sut tiu th ca on mch MP l : A.120O; 240WB.120 3 O; 240WC.120O;120 3 W D.120 3 O; 240 3 W Cu 4:Chomchinxoay chiu RLC ni tip, L =0,637H, C = 39,8F, t vohai u mch in pc biu thc u = 150 2 sin100tt (V) mch tiu th cng sut P = 90 W. in tr R trong mch c gi tr l A. 180B. 50C. 250D. 90 Cu 5(C. 2010): t in p u = 200cos100tt (V) vo hai u on mch gm mt bin tr R mc ni tip vi mt cun cm thun c t cm 1tH. iu chnh bin tr cng sut ta nhit trn bin tr t cc i, khi cng dng in hiu dng trong on mch bng A. 1 A.B. 2 A.C.2 A.D.22A. Cu 6: Ln lt mc vo ngun xoay chiu (200V-50Hz) :in tr thun,cun dy thun cm,t in th cng hiu dng ca dng in qua chng ln lt u bng 2A.Mc ni tip 3 phn t vo ngun xoay chiu trn th cng sut tiu th ca mch bng: A. 200WB. 400WC. 100WD. 800W DNG 10H S CNG SUT :. A. Phng php : cos = RZ.hay RUcosU =haycos.PU I =0 1 cos s sHoc dng gin vect b. ngha h s cng sut: b1 Tr.hp cos = 1:- Trong tr.hp ny =0: y l tr.hp on mch in xoay chiu ch cha R, hoc mch RLC nhng xy ra cng hng. Lc ny P=UI. R B C L A NM I UC UL UR O UAN UMB A B 26 b2. Tr.hp cos = 0: - = 2t. y l tr.hp on mch xoay chiu khng cha in tr thun.( on mch ch c L , hoc C hoc LC ni tipP = 0 b3. Tr.hp 0 < cos L= 14Ht Bi 6( H10-11): t in p xoay chiu c gi tr hiu dng v tn s khng i vo hai u on mch gm bin tr R mc ni tip vi t in c in dung C. Gi in p hiu dng gia hai u tu in, gia hai u bin tr v h s cng sut ca on mch khi bin tr c gi tr R1 ln lt l UC1, UR1 v cos1; khi bin tr c gi tr R2 th cc gi tr tng ng ni trn l UC2, UR2v cos2. Bit UC1 = 2UC2, UR2 = 2UR1. Gi tr ca cos1 v cos2 l: A. 1 21 2cos , cos3 5 = = .B. 1 21 1cos , cos5 3 = = . C. 1 21 2cos , cos5 5 = = .D. 1 21 1cos , cos2 2 2 = = . Hng dn gii :H s cng sut ca on mch tng ng vi hai gi tri ca R l: 2 2 2 2 2 2 1 21 2 1 1 2 22 2 2 21 1 2 2cos ; os ; (1); (2)R RR C R CR C R CU Uc U U U U UU U U U = = = + = ++ +U .t (1) v (2) v theogithittatmcUR1=UC1/2,thayvohaicngthctrnvhscngsut,tac 1 21 2os ' os5 5c c = =27 Hay :UC1 = 2UC2 I 1 = 2I 2 ( v C khng i), UR2 = 2UR1 I 2 R 2 = 2I 1 R 1 R 2 = 4R 1, Z C = 2 R 1

cos = RZ Chn C Bi 7( H10-11): Trong gi hc thc hnh, hc sinh mc ni tip mt qut in xoay chiu vi in tr R ri mc hai u on mch ny vo in p xoay chiu c gi tr hiu dng 380V. Bit qut ny c cc gi tr nh mc : 220V - 88W v khi hot ng ng cng sut nh mc th lch pha gia in p hai u qut v cng dng in qua n l , vi cos = 0,8. qut in ny chy ng cng sut nh mc th R bng A. 180 OB. 354OC. 361OD. 267O Hng dn gii :Ta c th xem qut nh mt cun dy c in tr r.Cng sut ca qut c xc nh theo cng thc:2cos 0, 5 , 352quatPP UI I A rI = = = = O.TngtrcamchgmqutvintrRlZ=U/I =760(m),suyra:cmkhngcacundycaqutcxcnhtheocng thc:21 ostan 264osLLc ZZ rr c= = = O.Vyintrcacundycxcnhtheocng thc:2 2 2( ) 361LZ R r Z R = + + ~ O Hay: Coi on mch cha qut gm cun dy v in tr r mc ni tip. Ta cgin vc t nh hnh v ( = QU ,RU ) +) Ta c phng trnh : U2 = UQ2 + UR2 + 2 UQURcos => UR = 180,33V +) PQ = UQ Icos => I = 0,5A+) R = UR/I= 361O Bi 11 (Cu 18 24 cc kho th)on mch xoay chiu gm cun dy v t in C mc ni tip . in p hiu dng 2 u on mch l U=120V .Bit h s cng sut on mch l 0,8 v h s cng sut cun dy l 0,6. Cho bit dng in tr pha so vi in p hai u on mch . in p hiu dng hai u cun dy v hai u t in ln lt l : A.80V; 60VB.90V; 30VC.128V; 72VD.160V; 56V HNG DN : V gin vect , ta thy : cos 0, 6RddUU = = , cos 0, 8RUU = =0, 6 30, 8 4RdRUUUU= = 34dUU= Ud=4.U/3=120.4/3=160V Ngoi ra ta thy : sin cos 0, 6 OBA = = 2 2cos (1 0, 6 ) 0, 64 OBA= = cos 0,8 OBA=Dng h thc lng trong tam gic :OAB2 2 22 . .cosd C CU U U UU OBA = + + ; th s tm UC .Hay :2 2 22. . .cosC d dU U U UU AOB = + +; vicos cos .cos sin .sind dAOB = Cu 13.(C 2011-12)t in p150 2 os100 u c t t = (V) vo hai u on mch gm in tr thun, cun cm thun v t in mc ni tip th in p hiu dng gia hai u in tr thun l 150 V. H s cng sut ca mch l A. 32.B. 1.C. 12.D. 33. HNG DN ::150cos 1150URU = = = RU

QUU

dUA Od H RU

CUUB

28 DNG 11 : XC DNH TN S DNG IN A. phng php : 1/ Dng cc cng thc c lin quan n tn s gc ,tn s : -E0= NBeS ; 22 f 2 nTte= = t = t ;ZC = 12 f .C=e t1C ; ; ZL = eL; 1LLtanRe e =-iu kin c cng hng l :1LCe= hay 2LC =1hay 2 2LC =1 4 f t

hay1L CZ Z LCee= =

-Lin h gia Z v tn s f : f0 l tn s lc cng hng . Khi f f0 : Mch c tnh cm khng , Z v f ng bin 2/ Mch RLC c e; f thay i:Lu : L v C mc lin tip nhau * Khi 1LCe =th IMax URmax; PMax cn ULCMin * Khi 1 122CL RCe = th 2 .ax2 24U LULMR LC RC= * Khi 212L RL Ce = th 2 .ax2 24U LUCMR LC RC= * Vi e = e1 hoc e = e2 th I hoc Phoc UR c cng mt gi tr th IMax hoc PMax hoc URMax khi 1 2e ee = tn s 1 2f f f =B. p dng ; Bi 1 : Gia hai bn t in c in p xoay chiu 220V, 60Hz. Dng in qua t in c cng 0,5A. dng in qua t in c cng bng 8A th tn s ca dng in l A. 15Hz.B. 240Hz.C. 480Hz.D. 960Hz. Hng dn: : 1 1 2 2C1 C22 2 221 1U UI 200.2 .f 0, 5A; I 200.2 .f 8A;Z ZI f f 8f 960Hz.I f 0, 5 60= = t = = = t = = = =

Bi 2 CB. Trong mch in xoay chiu gm R, L, C mc ni tip. Nu tng tn s ca in pxoay chiu hai u mch th: A. dung khng tng. B. cm khng gim. C. in tr R thay i.D. tng tr ca mch thay i. Bi 3: Mt cun dy dn c h s t cm L c mc ni tipvi mt t c in dung C ri mc vo 3 im A, B camt mch in xoay chiu c tn s f. o in p gia hai u on mch AB, gia hai u cun dy vgia hai cc ca t in bng vn k c in tr rt ln,ta ln lt c: UAB = 37,5 V, Ud=50V, UC =17,5 V. o cng dng in bng mt ampe k c in tr khng ng k, ta thy I=0,1 A.Khi tn s f thayi n gi tr fm=330 Hz th cng dng in trong mch t gi tr cc i.Tnh tn s f ca in p s dng trn. Hng dn gii : Gi s cun dy thun cm khng c in tr r th:UAB = UL UC = 50 17,5 = 32,5 V. Khng ph hp vi gi tr cho. Nn cun dy phi c in tr trong r ng k. Tng tr cun dy:2 2Z r Zd L= + C B L,r. A Z Dung khng Cm . .khngZmin Of0f

29 Bin v gi tr hiu dng ca cng dng din c tnh theo cc cng thc: 0 002 2U UIZr Zd ddL= =+v2 2U UIZr Zd ddL= =+ Cng sut tiu th ca cun dy: P = Ud.I.cosd = I.r2 Vi h s cng sut: cosd= 2 2r rZZ r dL=+ Ta tnh c:Tng tr ca cun dy: U 50Z 500I 0,1dd = = = O Dung khng ca t in: U 17, 5Z 175I 0,1CC = = = O.Tng tr ca on mch: U 37, 5Z 375I 0,1ABAB = = = O Khi f = fm, trong mch c cng hng (Imax) nn:em2 = 1LC 2 2 21 1 1LC=(2 f ) (2. .330)m me t t = = (1) Mt khc:ZAB2 = r2 + (ZL ZC.2 = r2 + ZL2 2ZLZC + ZC2 ;ZAB2 = Zd2 + ZC2 2ZLZC 2ZLZC = Zd2 + ZC2 ZAB2 = 5002 + 1752 - 3752 = 14.104

2.L. e .1C.e=4 4 4L L2 14.10 7.10 L=7.10 .CC C= = (2) Th (2) vo (1) ta c: 7.104.C2 = 21(2. .330) t Suy ra: C=1,82.10-6 F; L=7.104.C=7.104.1,82.10-6=0,128HM: ZC = 1C.e=61 1 1f= 500C.2. f C.2. .Z 1,82.10 .2.3,14.175ct t = = Hz Bi 4( 21 cc kho th )Mch in xoay chiu R,C, cun dy c Lni tip vi L khng i ;cn R thay i . t 1LCe = Cn phi t vo hai u on mch 1 in p xoay chiu c gi tr hiu dng khng i nhnge thay i .Hie bng bao nhiu in p hai u cun dy khng ph thuc vo R ? A.0e e = B.02 e e = C.02. e e = D.012e e =Bi 5 ( 23 cc kho th )Mch in gm ba phn t R1,L1,C1 c tn s cng hng e1 v mch in gm ba phn t R2,L2,C2 c tn s cng hng e2 (1 2e e = ). Mc ni tip 2 mch vi nhau th tn s cng hng l e l : A.1 2. e e e =B.2 21 1 2 21 2L LL Le ee+=+C.1 2. e ee = D.2 21 1 2 21 2L LC Ce ee+=+ Bi 6(Cu 44 24 cc kho th) Cho on mch RLC ni tip ; R=30O; 0, 4( ) L Ht= ; 410( ) C Ft= . Khie thay i th thy khie bin thin t50 ( / ) rad s t n150 ( / ) rad s t , cng dng in trong mch s : A.tngB.gimC.tng ri sau gimD.gim ri sau tng HNG DN tnh 01LCe=khi c cng hng .. thy 0e nm trong khong gia 50 ( / ) rad s t v 150 ( / ) rad s tsuy ra I lc u tng n gi tr Imax ri sau I gimBi 7( H 2011-2012) : t in p xoay chiu u = U0coset (U0 khng i v e thay i C. vo hai u on mch gm in tr thun R, cun cm thun c t cm L v t in c in dung C mc ni tip, vi CR2 < 2L. Khi e = e1 hoc e = e2 th in p hiu dng gia hai bn t in c cng mt gi tr. Khi e = e0 th in p hiu dng gia hai bn t in t cc i. H thc lin h gia e1, e2 v e0 l A. 0 1 21( )2e= e+e B. 2 2 20 1 21( )2e= e+e C. 0 1 2e= ee D. 2 2 20 1 21 1 1 1( )2= +e e e HNG DN ::I Ur Ud UL d

dUA OH RU

CUUB

30 +Theo bi cho : e e e e = = =1 2 1 2hoc t hU .C CUv e ee ee e=+ + 1 22 2 2 21 21 21 1 y :1 1( ) ( )C CKhiR L R LC Cee eeee+ =+ 2 21 22 122 21221( ) 1( )R LCR LC Bin i my dng thu c : 2 22 2 2 2 2 21 2 1 2 2 22 2 1( ) ( ) 2( )(1)2L R RL RC CL L LC Le e e e + = + = = + Mt khc, khiebin thin c UCmax th : 220 21(2)2RLC Le = T (1) v (2) suy ra p n : 2 21 2 201 1( )2e ee= + Bi 8 ( H 2011-2012) : Ln lt t cc in p xoay chiu u1 = 12cos(100 ) U t t + ; u2 =22cos(120 ) U t t +v u3 =32cos(110 ) U t t +vo hai u on mch gm in tr thun R, cun cm thun c t cm L v t in c in dung C mc ni tip th cng dng in trong on mch c biu thc tng ng l: i1 =2 cos100 I t t ; i2 = 22 cos(120 )3I ttt +v i3 = 2' 2 cos(110 )3I ttt . So snh I v I, ta c: A. I = I.B. I =' 2 I .C. I < I.D. I > I. Gii :. Da vo th hnh bn ta thy i vi on mch xoay chiu c R,L,C ni tip ; khi cho f thay i th I thay i ;Khi f = f1 hoc f = f2 th I c cng mt gi tr th IMax khi tn s 0 1 2f f f = .Bi cho f1=50Hz ,f2=60Hz,f3=55Hz v cho khi f = f1=50Hz hoc f = f2 =60Hz th I1 =I2=I . Suy ra : 0 1 250.60 54, 77 f f f Hz = = =. y cho khi f=f3 th cng l I. Qua th , ta thy f1 >f3>f2v f3 ~f0 . Suy ra I~Imax nn I >I VN DNG : Cu1:Mtmch R,L,C mc ni tip trong R = 120 O, L =2/ t H vC = 2.10-4/ t F, ngun c tn s f thay i c. i sm pha hn u, f cn tho mnA: f> 12,5HzB: fs 12,5HzC. f< 12,5Hz D. f < 25Hz Cu 2. in p t vo hai u t in l U = 110 V, tn s f1 = 50Hz. Khi dng in qua t l I1 = 0,2A. dng in qua t l I2 = 0,5 A th cn tng hay gim tn s bao nhiu ln? A. 5 ln B. 3,5 lnC. 3 ln D. 2,5 ln. Cu 3(DH 2009):t in p xoay chiu u = U0coset c U0 khng i v e thay i c vo hai u on mch c R, L, C mc ni tip. Thay i e th cng dng in hiu dng trong mch khi e = e1 bng cng dng in hiu dng trong mch khi e = e2. H thc ng l : A. 1 22LCe+ e= .B. 1 21.LCee= .C. 1 22LCe+ e= .D. 1 21.LCee= . (HNG DN ::( ) ( )( )1 2 1 1 2 2 1 1 2 22 22 21 1 2 21 21 2 1 21 1 1 1 1. ....L C L C L C C LL C L CU UI I Z Z Z Z Z Z Z ZR Z Z R Z ZL LC Ce ee e ee= = = = + + | |+ = + = |\ .Cu 4: Mt mch R,L,C mc ni tip trong R = 120 O, L = 2/ t H v C = 2.10 - 4/ t F, ngun c tn s f thay i c. i sm pha hn u, f cn tho mnA: f> 12,5Hz B: fs 12,5HzC. f< 12,5HzD. f < 25Hz I=I1=I2 f1f2 f O f0 I I max 31 Cu 5 ( 21 cc kho th )t in p 0cos ( ) u U t V e = c tn s gc thay i vo hai u on mch RLC mc nitip . Khi tn s gc l 100 / rad s thoc25 / rad s tth cng dng in hiu dng qua mch bng nhau . cng dng in hiu dng qua mch cc i th tn s gc bng : A. 60 / rad s t B. 55 / rad s t C. 45 / rad s t D. 50 / rad s tHNG DN :* Vi e = e1 hoc e = e2 th I hoc Phoc UR c cng mt gi tr th IMax hoc PMax hoc URMax khi 1 2e ee =.p dng ta c :100 .25 50 / rad s e t t t = =Cu 6: Cho mch xoay chiu khng phn nhnh RLC c tn s dng in thay i c. Gi 2 1 0; ; f f fln lt l cc gi tr ca tn s dng in lm cho max max max; ;C L RU U U . Ta c A. 2001ffff=B. 2 1 0f f f + = C. 210fff = D. 0 1 2f f f = Cu 7Mt on mch in gm t in c in dung 10-4/t F mc ni tip vi in tr 125 O, mc on mch vo mng in xoay chiu c tn s f. Tn s f phi bng bao nhiu dng in lch pha t/4 so vi in p hai u mch. A. f = 50\3 HzB. f = 40 HzC. f = 50HzD. f = 60Hz 1221 40fLfCHD: tan f HzRtt= = = DNG 9 BI TON CC TR A. Phng php : 1.on mch RLC c R thay i: * Khi R=ZL-ZC th 2 2ax22U UPMRZ ZL C= =; =R2 Ucoskhi oU =22 * Khi R=R1 hoc R=R2 th P c cng gi tr. Ta c 22; ( )1 2 1 2UR R R R Z ZL CP+ = = V khi 1 2R R R =th 2max21 2UPR R=* Tr.hp cun dy c in tr R0 (hnh v) Gi PM l cng sut tiu th in trn ton mch ;PR l cng sut tiu th in trnbin tr R: Khi 2 2ax 02( )20U UR Z Z R PL Mm CR RZ ZL C= = =+ Khi 2 22 2( )ax 02( ) 2 22 ( ) 2 00 0U UR R Z ZL Rm CR RR Z Z RL C= + = =++ +P2. on mch RLC c L thay i: Lu : L v C mc lin tip nhau * Khi 21LC e=th IMax URmax; PMax cn ULCMin * Khi 2 2R ZCZLZC+=th 2 2axU R ZCULMR+= v2 2 2 2 2 2ax ax ax; 0LM R C LM C LMU U U U U UU U = + + = * Vi L = L1 hoc L = L2 th UL c cng gi tr th ULmax khi 21 1 1 11 2( )21 21 2L LLZ Z Z L LL L L= + =+ * Khi 2 242Z R ZC CZL+ +=th 2 Rax2 24UURLMR Z ZC C=+ Lu : R v L mc lin tip nhau A B C RL,R0 32 3. on mch RLC c C thay i: *Suy ra +=2 2R ZLZCZL v+=2 2( )U R ZLUCMaxR * Khi 21CL e=th IMax URmax; PMax cn ULCMinLu : L v C mc lin tip nhau * Khi 2 2R ZLZCZL+=th 2 2axU R ZLUCMR+=v2 2 2 2 2 2; 0ax ax axU U U U U UU UR L L CM CM CM= + + =* Khi C = C1 hoc C = C2 th UC c cng gi tr th UCmax khi 1 1 1 11 2( )2 21 2C CCZ Z ZC C C+= + = * Khi 2 242Z R ZL LZC+ +=th 2 Rax2 24UURCMR Z ZL L=+ Lu : R v C mc lin tip nhau 4. Mch RLC c e ; f thay i: * Khi 1LCe =th IMax URmax; PMax cn ULCMinLu : L v C mc lin tip nhau * Khi 1 122CL RCe = th 2 .ax2 24U LULMR LC RC= * Khi 212L RL Ce = th 2 .ax2 24U LUCMR LC RC= * Vi e = e1 hoc e = e2 th I hoc Phoc UR c cng mt gi tr th IMax hoc PMax hoc URMax khi 1 2e ee = tn s 1 2f f f =Lu : i lc ta c th gii bi ton bng gin vect B.p dng : Bi 1 : TNPT ( 2001) Cho : R thay i t 0 n vi trmO; C = 410 Ft 50 2cos100 ( )ABu t V t = a) iu chnh cho R = 75O.Tnh Z ? UC ? B. Dch chuyn con chy v bn phi .Cng sut ta nhit ca mch thay i nh th no ? Tnh Pmax ? HNG DN : : a) Z = 125O; UC = IZC = 40V B. 22.1.CUP I RR ZR= =+ 2 2max CP R Z = 100CR Z = = O Pmax = 212, 52UWR =Khi R tng th Pmax =12,5W sau gim xung khi R tip tc tng n vi trmO Bi 2:Cho mch RLC c R=100 O ; C4102Ft= cun dy thun cm c L thay i c. t vo Hai u mch in p100 2 os100 t(V) u c t =Tnh L ULC cc tiu A. 1L Ht= B. 2L Ht= C. 1,5L Ht= D. 210L Ht=HNG DN :: L min2Z2R12(Z )LLC LC C LCU UU Z U Z LZZCt= = = =+ CR A 33 Bi3:MtonmchgmintrRnitipvicunthuncmLvtxoayC.BitR=100O, L=0,318H. t vo 2 u on mch mt in p u=200 2 cos 100tt (V).Tm in dung C in p gia 2 bn t in t gi tr cc i. Tnh gi tr cc i . Hng dn gii :Cc tr lin quan n in p cc i : -Khi L thay i, C vfkhng i UL cc i th2 2CLCR ZZZ+= . vi ULmax = 2 2CUR ZR+ . -Khi C thay i, L vfkhng i UC cc i th2 2LCLR ZZZ+= . vi UCmax =.2 2LUR ZR+-Khi tn s f thay i cn L v C khng i UC cc i th 2 222 222LC R CCLe= . -Ta c th dng o hm : ZL=eL=100O -in p gia 2 bn t in :yUZZZZUZ Z Z Z RZ UZ I UCLCL C C L LCC C=+ +=+ += =12 R . 2..22 2 2 2 2 -UC max khi y = y min m y l hm parabol vi i s l CZx1=-vy y min khi 2 21LLCZ RZZx+= =(nh parabol) O =+= =+= 20012 22 2minLLCLZZ RxkhiZZ RRyvyF Ct 2104 =v UC max = 200 2(V) Bi 4:Cho mch in xoay chiu RLC c: R=100 O ; L=2Ht, in dung C ca t in bin thin. t vo hai u mch in p200 2 os100 t(V) u c t = . Tnh C in p gia hai u t in t gi tr cc i A 4102C Ft= B.4102.5C Ft= C. 4104C Ft=D. 2102C Ft= HNG DN : Gii:CHN B : UCmax khi2 2LCLR ZZZ+=Bi 5:Cho mch RLC c R=100 O ; C4102Ft= cun dy thun cm c L thay i c. t vo Hai u mch in p100 2 os100 t(V) u c t =Tnh L ULC cc tiu A. 1L Ht= B. 2L Ht= C. 1,5L Ht= D. 210L Ht=HNG DN :: 2ZL min2R12(Z )LU UU Z U Z LLC LC C LCZZCt= = = =+ Bi 6:Cho mch in xoay chiu nh hnh v ,200cos100 ( )ABu t V t = , t c in dung ) (. 2104F Ct=, cun dy thun cm c t cm 1( ) L Ht=, R bin i c t 0 n 200O.Tnh R cng sut tiu th P ca mch cc i. Tnh cng sut cc i . A.100WB.200W C.50WD.250W Hng dn gii :Chn A.+Cng sut tiu th trn mch c bin tr R ca on mch RLC cc i khi R = |ZL ZC| v cng sut cc i l Pmax = | | . 22C LZ ZU. Bi 7 (H 2008): on mch in xoay chiu gm bin tr R, cun dy thun cm c t cm L v t in cindungCmc ni tip.Bit in p hiudnghai u on mch lU, cm khngZL, dung khng ZC (vi ZC = ZL) v tn s dng in trong mch khng i. Thay i R n gi tr R0 th cng sut tiu th ca on mch t gi tr cc i Pm, khi C AB RL NM 34 A. R0 = ZL + ZC.B. 2m0UP .R= C. 2LmCZP .Z= D. 0 L CR Z Z = HNGDN:: ( )2222L CU RP = I R =R + Z - Z( )2max 2L CUZ - ZR +RP =( )2L CZ - ZR =Rkhi0 L C2max0R = Z - ZUP =2R Bi 8( H 2009): : t in p xoay chiu c gi tr hiu dng 120V, tn s 50 Hz vo hai u on mch mc ni tip gm in tr thun 30 O, cun cm thun c t cm 0, 4t(H) v t in c in dung thay i c. iu chnh in dung ca t in th in p hiu dng gia hai u cun cm t gi tr cc i bng A. 150 V.B. 160 V.C. 100 V.D. 250 V. HNG DN ::. .40 . ; .L LL LMAX MAX LMINU Z U ZZ H S U I ZZ R= O = = = = = 120.40/30=160V cng hng in..) Bi9 ( H 2011-2012) : t in p xoay chiu u = U0coset (U0 khng i v e thay i C. vo hai u on mch gm in tr thun R, cun cm thun c t cm L v t in c in dung C mc ni tip, vi CR2 < 2L. Khi e = e1 hoc e = e2 th in p hiu dng gia hai bn t in c cng mt gi tr. Khi e = e0 th in p hiu dng gia hai bn t in t cc i. H thc lin h gia e1, e2 v e0 l A. 0 1 21( )2e= e+e B. 2 2 20 1 21( )2e= e+e C. 0 1 2e= ee D. 2 2 20 1 21 1 1 1( )2= +e e e HNG DN ::+Theo bi cho : e e e e = = =1 2 1 2hoc t hU .C CUv e ee ee e=+ + 1 22 2 2 21 21 21 1 y :1 1( ) ( )C CKhiR L R LC Cee eeee+ =+ 2 21 22 122 21221( ) 1( )R LCR LC Bin i my dng thu c : 2 22 2 2 2 2 21 2 1 2 2 22 2 1( ) ( ) 2( )(1)2L R RL RC CL L LC Le e e e + = + = = + Mt khc, khiebin thin c UCmax th : 220 21(2)2RLC Le = T (1) v (2) suy ra p n : 2 21 2 201 1( )2e ee= + Bi 10( H 2011-2012) : t in p u =2 cos 2 U ft t(U khng i, tn s f thay i C. vo hai u on mch mc ni tip gm in tr thun R, cun cm thun c t cm L v t in c in dung C. Khi tn s l f1 th cm khng v dung khng ca on mch c gi tr ln lt l 6O v 8 O. Khi tn s l f2 th h s cng sut ca on mch bng 1. H thc lin h gia f1 v f2 lA. f2 = 12.3f B. f2 = 13.2f C. f2 = 13.4 f D. f2 = 14.3 fHNG DN : Khi tn s f1 : 2 111 16 3 18 2 2LCZLC LCZ fet= = = khi tn s f2 = 13221fLC=t VN DNGCu 1: Cho mch mc theo th t RLC mc ni tip, t v hai u on mch in p xoay chiu ,bit R v L khng i cho C thay i .Khi UC t gi tr cc ith h thc no sau y l ng A. U2Cmax= U2 + U2(RL)B. UCmax = UR + UL C. UCmax= UL2 D. UCmax =3 UR. Cu 2: Cho mch xoay chiu khng phn nhnh RLC c tn s dng in thay i c. Gi 2 1 0; ; f f fln lt l cc gi tr ca tn s dng in lm cho max max max; ;C L RU U U . Ta c 35 A. 2001ffff=B. 2 1 0f f f + = C. 210fff = D. 0 1 2f f f = Cu 3(C. 2010): t in p u =U 2 cos t e(V) vo hai u on mch gm cun cm thun mc ni tip vi mt bin tr R. ng vi hai gi tr R1 = 20 O v R2 = 80 O ca bin tr th cng sut tiu th trong on mch u bng 400 W. Gi tr ca U l A. 400 V.B. 200 V.C. 100 V.D. 100 2 V. Cu 2. Cho mch in nh hnh v .t vo hai u on mch mt in pxoay chiu n nh c in phiu dng U .Thay i R n gi tr R0 th cng sut on mch cc i .Tm cng sut cc i . A.2max02UPR=B.2max0UPR= C. 2max02UPR=D)2max04UPR=Cu 4. Mt mch in RLC ni tip, R l bin tr, in p hai u mch) V ( t 100 cos 2 10 u t = . Khi iu chnh R1 =O 9 v R2 =O 16th mch tiu th cng mt cng sut. Gi tr cng sut l:A. 8WB.2 4 , 0 W C. 0,8 W D. 4 W Cu 5 t mt in p xoay chiu vo hai u on mch RLC ni tip c R thay i th thy khi R=30O v R=120O th cng sut to nhit trn on mch khng i. cng sut t cc i th gi tr R phi l A. 150OB. 24O C. 90OD. 60O Cu 6Mt mch R,L,C mc ni tip m L,C khng i R bin thin. t vo hai u mch mt ngun xoay chiu ri iu chnh R n khi Pmax, lc lch pha gia U v I l A: 6t B: 3tC.4tD. 2t Cu7:Chomchinnhhnhv, 0,6Lt= (H), 410Ct= (F),r=30(O),uAB=100 2 cos100tt(V). Cng sut trn R ln nht khi R c gi tr: A. 40(O)B. 50(O)C. 30(O)D. 20(O) Cu 8: Cho on mch RLC nh hnh v, uAB = 100 2 cos100tt(V). Thay i R n R0 th Pmax = 200(W). Gi tr R0 bng: A. 75(O)B. 50(O)C. 25(O)D. 100(O) Cu 9: Cho mch in RLC ni tip, trong cun L thun cm, R l bin tr .in phiu dng U=200V, f=50Hz, bit ZL = 2ZC,iu chnh R cng sut ca h t gi tr ln nht th dng in trong mch c gi tr l I= . Gi tr ca C, L l: A.110mtF v2HtB. 310tmF v 4HtC. 110tF v 2mHt D. 110tmF v 4Ht Cu 10: Cho on mch R,L,C trong L bin thin c , R = 100O, in phai u on mch 200 os100 t ( V) u c t = . Khi thay i L th cng dng in hiu dng t gi tr cc i lA. 2A. B. 0,5 A. C. 12 AD.2 A. DNG 10 LCH PHA CA IN P CA 2 ON MCH MC NI TIP A.Phng php : 1.HaionmchAMgmR1L1C1nitipvonmchMBgmR2L2C2nitipmcnitipvi nhau Nuc UAB = UAM + UMB uAB; uAM v uMB cng pha tanuAB = tanuAM = tanuMB 2. Hai on mch R1L1C1 v R2L2C2mc ni tipc pha lch nhau A Vi 1 1tan11= Z ZL CR v 2 2tan22= Z ZL CR (gi s 1 > 2) C 1 2 = A 1 21 2tan tantan1 tan tan = A+ Tr.hp c bit A = t/2 (vung pha nhau) th tan1tan2 = -1.VD: * Mch in hnh 1 c uAB v uAM lch pha nhau A y 2 on mch AB v AM c cng i v uAB chm pha hn uAM R B C r, L A R B C L A RLCMAB Hnh 1 ARL C B 36 RLCMAB Hnh 2 RLCMAB AM AB = A tan tantan1 tan tan = A+AM ABAM AB Nu uAB vung pha vi uAM thtan tan =-11Z Z ZL C LAM ABR R = * Mch in hnh 2: Khi C = C1 v C = C2 (gi s C1 > C2) th i1 v i2 lch pha nhau A yhai on mch RLC1 v RLC2 c cng uAB Gi 1 v 2 l lch pha ca uAB so vi i1 v i2

th c 1 > 2 1 - 2 = A Nu I1 = I2 th 1 = -2 = A/2 Nu I1 = I2 th tnh 1 21 2tan tantan1 tan tan = A+ 3. Lin quan lch pha: a. Tr.hp : 1 2 1 2tan .tan 12t + = = b. Tr.hp : 1 2 1 2tan .tan 12t = = c. Tr.hp : 1 2 1 2tan .tan 12t + = = 4. Xt on mch AB nh hnh v (1) Nu : AM AB = A (2a) tan tantan1 tan tan = A+AM ABAM AB tan 1L C LL C LZ Z ZR RZ Z ZR R= A+ hay 2tan( )CL L CRZR Z Z Z = A+ (2b) B. p dngBi 1 ( Bi 2Trang 174 NC ) Cho mch in nh hnh v 174:Cho R = 100O; UR = 50V;UL = 50V ; UC = 87,5V ; f = 50Hz a) Tnh L ? C ?b). Tnh Z ? UAB ? c). Tnh lch pha ca uAN v uMB ? lch pha ny l :2 tthR phi bng bao nhiu ? ( L , C ,fkhng i ) HNG DN : : a) ZL = 100; ZC = 175 . b). Z = 125 ; UAB = 62,5 ( V ) c) 4ANt = ; 2MNt = ,34 2 4t t t A = + = , : 2t A =( ). 1Z ZC LR R = R =. 50 7 132, 3 Z ZL C= ~ O Bi 2(H 2008): Cho on mch in xoay chiu gm cun dy mc ni tip vi t in. lch pha ca in p gia hai u cun dy so vi cng dng in trong mch l 3t. in p hiu dng gia hai u t in bng3ln in p hiu dng gia hai u cun dy. lch pha ca in pgia hai u cun dy so vi in p gia hai u on mch trn l A. 0.B. 2t.C. 3t .D. 23t. HNG DN :: LcdZ tan = = tan = 3r 3( )2 2 2 2 2C L r C LU = 3. U + U Z = 3 Z +r LCZ = 3.rZ = 2 3.r L CZ - Ztan = = - 3r = -323cdt =CRL - D A B B (3) 37 Bi3(H2008):ChoonmchinxoaychiugmcundycintrthunR,mcnitipvit in. Bit in p gia hai u cun dy lch pha 2t so vi in p gia hai u on mch. Mi lin h gia in tr thun R vi cm khng ZL ca cun dy v dung khng ZC ca t in l A. R2 = ZC(ZL ZC..B. R2 = ZC(ZC ZL).C. R2 = ZL(ZC ZL).D. R2 = ZL(ZL ZC.. HNG DN :: ( ) = = = L C Lcd L C LZ Z Ztan .tan . R Z Z ZR R21Bi 4( H10-11): Mt on mch AB gm hai on mch AM v MB mc ni tip. on mch AM c in tr thun 50O mc ni tip vi cun cm thun c t cm 1tH, on mch MB ch c t in vi in dung thay i c. t in p u = U0cos100tt (V) vo hai u on mch AB. iu chnh in dung ca t in n gi tr C1 sao cho in p hai u on mch AB lch pha 2t so vi in p hai u on mch AM. Gi tr ca C1 bng A. 54.10FtB. 58.10FtC. 52.10FtD. 510Ft Hng dn gii : 1. 1L C LZ Z ZR R= ZC1=125OC1=58.10Ft Hay : lch pha gia in phai u on mch AN v i l : tan (1)LAMZR = . lch pha gia u v I l 1tanL CZ ZR= (2).Theo gi thit th 2 511 1 2( ) 8.10tan tan 1 1 1252L L CAM AM C LLZ Z Z RZ Z C FR Zt t+ = = = = + = O =Bi 5( 23 cc kho th )Mch in xoay chiu gm cun dy c in tr thun R mc ni tip vi t in c in dung C . Bit mi lin h gia R ,ZL,ZC l R2=ZL.(ZC-ZL). in p hai u cun dy lch pha so vi din p hai u on mch gc : A.4tB.2tC.3tD.6t VN DNG Cu1:ChomchinxoaychiuRLCnhhnhv ( ) V ft U uABt 2 cos 2 = .Cundythuncmc tcmH Lt 35= ,tdincF Ct 24103 = .HtuNB v uAB lch pha nhau 900 .Tn s f ca dng in xoay chiu c gi tr l A. 120HzB. 60HzC. 100HzD. 50H Cu 2: Mt on mch in xoay chiu c dng nhhnh v.Bit in puAE v uEB lch pha nhau 900.Tmmi lin h gia R,r,L,.C.A. R = C.r.LB. r =C. R..L C. L = C.R.r D.C = L.R.r Cu 3: Mch in AB gm hai on mch AM (cha R1 ni tip L = 2H) v MB (cha R2 ni tip C = 100F) mc ni tip nhau. t vo hai im A, B mt in pu = 200 2 cos(100t)V th o c in pgia hai im A, M bng 120V v gia hai im B, M bng 160V. R1 v R2 tha mn iu kin A. R1/R2 = 2.B. R1.R2 = 200 2.C. R1.R2 = 2.104 2.qD. R2/R1 = 2. Cu 4(C. 2010): t in p220 2 cos100 u t t =(V) vo hai u on mch AB gm hai on mch AM v MB mc ni tip. on AM gm in tr thun R mc ni tip vi cun cm thun L, on MB ch c t in C. Bit in p gia hai u on mch AM v in p gia hai u on mch MB c gi tr hiu dng bng nhau nhng lch pha nhau 23t. in p hiu dng gia hai u on mch AM bng A.220 2 V.B. 2203V.C. 220 V.D. 110 V.

RL

C

ABN A C REL, r B 38 Cu 5: C 2 cun dy mc ni tip vi nhau,cun 1 c t cm 1L ,in tr thun 1R ,cun 2 c t cm 2L ,in tr thun 2R .Bit1L2R = 2L1R. in p tc thi 2 u ca 2 cun dy lch pha nhau 1 gc: A.t /3B.t /6C.t /4D. 0 Cu 6: Hai cun dy (R1,L1) v (R2,L2)`mc ni tip nhau v t vo mt `in p xoay chiu c gi tr hiu dng U. Gi U1 v U2 l `in p hiu dng tng ng gia hai cun (R1,L1) v (R2,L2). iu kin U=U1+U2 l: A. 1 22 1L LR R= ; B. L1L2=R1R2; C. L1+L2=R1+R2 D. 2211RLRL= ; Cu 7 : Cho on mch nh hnh v : Bit R1 = 4O, C1 = 2108tF, R2 = 100O, L = 1tH, f = 50Hz,C2 l t bin i. Thay i C2 in p uAE

cng pha vi uEB. Gi tr C2 khi l :A.4103FtB.2103FtC.3103Ft D.1103Ft Dng 14 Bitu(t) hoc i(t)Tm u(t+nT/3) hoc Tmu(t+nT/4) hoc Tm i(t+nT/3) hoc Tm i(t+nT/4) A.Phng php : Dng mi lin quan gia dao ng iu ha v chuyn ng trn u . V vng trn lng gic c trc l u(t) hoc i(t) .Xc nh im chuyn ng trn u c hnh chiu u(t) hoc i(t) . Thm nT/3 hoc nT/4 vt chuyn thm gc n./3 hoc n./4. Xc nh v tr mi xong chiu xung trc tm gi tr u(t+nT/3) hoc tm u(t+nT/4) hoc tm i(t+nT/3) hoc tm i(t+nT/4) bng lng gic B.p dng : Cu 76 Biu thc in p hai u mt on mch: u = 200 cose t (V). Ti thi im t, in p u = 100(V) v ang tng. Hi vo thi im( t + T/4 ), in p u bng bao nhiu? A. 100 V. B. 100 2V. C. 100 3V.D. -100 V. HUNG DN Dng mi lin h gia dddh v chuyn ng trn u gii : T/4 th chuyn ng trn u quy vng v vy nu vo thi im t CDTD Qv ang tng thvo thi im ( t + T/4 ) CDTD M nh hnh v . Ta c : 100 2 2cos200 2o = = 4to = ; gc 4 MOP t = . Hnh chiu ca M xung Ox cho tau100 2V =.Cu 33. Cng dng in tc thi chy qua mt on mch in xoay chiu l ( ) 4sin 20 ( ) i t A t = , t o bng giy. Ti thi im 1tno dng in ang gim v c cng bng 22 i A = . Hi n thi im ( )2 10, 025 t t s = +cng dng in bng bao nhiu ? A.2 3A B.2 3A C.2AD.2A Cu 51: in p gia hai u mt on mch c biu thc) (2100 cos 2 220 V t u|.|

\| =tt ,ttnh bng giy (s). Ti mt thi im) (1s tno in p ang gim v c gi tr tc thi l) ( 2 110 V . Hi vo thi im ) ( 005 , 0 ) ( ) (1 2s s t s t + =th in p c gi tr tc thi bng bao nhiu ? A.) ( 3 110 V .B.) ( 3 110 V + .C.) ( 6 110 V .D.) ( 6 110 V + . DNG 15 GII TON BNG GIN VECT A. Phng php : 1) Cc quy tc cng vc t a) Quy tc tam gic Ni dung ca quy tc tam gic l: T im A tu ta v vc ta AB= , ri t im B ta v vc tb BC= . Khi vc tACc gi l tng ca hai vc t b va(Xem hnh a) . B. Quy tc hnh bnh hnh R1 L, R2 C2 B / / A C1 - E Q D + U0 02U P (C) x O M aBb a CB b a C A hnh a AbD hnh b 39 R C B L AMN hnh a.C R B L A M N Hnh b Ni dung ca quy tc hnh bnh hnh l: T im A tu ta v hai vc tb AD a AB= = v, sau dng im C sao cho ABCD l hnh bnh hnh th vc tACc gi l tng ca hai vc t b va(xem hnh B. . Ta thy khi dng quy tc hnh bnh hnh cc vc t u c chung mt gc A nn gi l cc vc t buc. Vn dng quy tc hnh bnh hnh cng cc vc t trong bi ton in xoay chiu ta c phng php vc t buc, cn nu vn dng quy tc tam gic th ta c phng php vc t trt (cc vc tni ui nhau). 2) C s vt l ca phng php gin vc t Xt mch in nh hnh a hoc hnh b. t vo 2 u on AB mt in pxoay chiu. Ti mt thi im bt k, cng dng in mi ch trn mch in l nh nhau. Nu cng dng in c biu thc l: ( )0cos i I t A e =th biu thc in p gia hai im AM, MN v NB ln lt l:( ) 2 cos2AM Lu U t Vte| |= + |\ .; ( ) 2 cosMN Ru U t V e =; ( ) 2 cos2NB Cu U t Vte| |= |\ .. + Do : NB MN AM ABu u u u + + = .+ Cc i lng bin thin iu ho cng tn s nn chng c th biu din bng cc vc t Frexnel: vc t.C R L ABU U U U + + =hoc AB AM MN NBU U U U = + +(trong ln ca cc vc t biu th in p hiu dng ca n). Cch v gin vc t cng gc :V i khng i nn ta chn trc cng dng in lm trc gc, gc ti im O, chiu dng l chiu quay lng gic.( H.1) *Chn trc ngang l trc cng d in *Chn gc A *V cc vecto ni dui , hoc v cngchung gc O( l A) + thc hin cng cc vc t trn ta phi vn dng mt trong hai quy tc cng vc t. Cch v gin vc t trt (Xem hnh ). Bc 1: Chn trc nm ngang l trc dng in, im u mch lm gc ( l im A). Bc 2: Biu din ln lt in pqua mi phn bng cc vc t MN NB; U ;AMU Uni ui nhau theo nguyn tc: R - i ngang; L - i ln; C - i xung. Bc 3: Ni A vi B th vc tABbiu din in p uAB. Tng t, vc t AN biu din in p uAN, vc tMB biu din in p uNB. Vc tABULURUABOU + L UCUCi+UABi+UANULUCURAMBNM RU N CUM RU N CULUUABB

AB hnh c A I + L UR UC US P x O UO UL UC ULC UR UIHnh ahnh b R C B L AMN 40 C R B L,r A M NHnh ar U MRUN CULUUABB

AB hnh c A I chnh l biu din uAB **Mt s im cn lu : + Cc in ptrn cc phn t c biu din bi cc vect m ln ca cc vect t l vi in phiu dng ca n. + lch pha gia cc in pl gc hp bi gia cc vecto tng ng biu din chng. lch pha gia in pv cng dng in l gc hp bi vecto biu din n vi trc I. Vc t nm trn (hng ln trn) s nhanh pha hn vc t nm di (hng xung di). + di cnh ca tam gic trn gin biu th in phiu dng, ln gc biu th lch pha. + Cng hai vc t cng phng ngc chiu 1 2; U Uthnh Utheo quy tc hnh bnh hnh (xem hnh trn). + Nu cun dy khng thun cm (trn on AM c c L v r (Xem hnh a di y)) thC R r L ABU U U U U + + + =ta v L trc nh sau: L - i ln, r - i ngang, R - i ngang v C - i xung (xem hnh B. hoc v r trc nh sau: r - i ngang, L - i ln, R - i ngang v C- i xung (Xem hnh c ) + Nu mch in c nhiu phn t (Xem hnh a) th ta cng v c gin mt cch n ginnh phng php nu (Xem hnh d).+ Gc hp bi hai vec tb val gc BAD(nh hn 1800). Vic gii cc bi ton l nhm xc nh ln cc cnh v cc gc ca cc tam gic hoc t gic, nh cc h thc lng trong tam gic vung, cc h thc lng gic, cc nh l hm s sin, hm s cos v cc cng thc ton hc. + Trong ton hc mt tam gic s gii c nu bit trc 3 (hai cnh mt gc, hai gc mt cnh, ba cnh) trong s 6 yu (ba gc trong v ba cnh). lm iu ta s dng cc nh l hm s sin v nh l hm s cosin (xem hnh bn). Cc cng thc thng dng cho tam gic vung: 2 2 2sin sin sin2 .cosa b cA B Ca b c bc A= == + .v 2 2 22 2 22 .cos2 .cosb c a ca Bc a b ab C = + = + mt s h thc trong tam gic vung:2 2 2a b c = +v2 2 21 1 1h b c= +v2'. ' h bc =Tm trn gin vct tam gic bit trc ba yu t (hai cnh mt gc, hai gc mt cnh), sau gii tam gic tm cc yu t cha bit, c tip tc nh vy cho cc tam gic cn li. B.p dng : Bi 1: R, L, C ni tip vi :R= 100W; UR =50V; UL=50V;UC=100V; f =50Hz a) L, C?b) ZAB? UAB? c) u uMB AN ? Da vo gin tm UAB Gii : a) I = U50R0,5AR 100= =; ZL= U50L100I 0, 5= = O Z100 1LL H100= = =e t t + ZC= U100C200I 0, 5= = O 41 1 10C FZ 100 .200 2C= = =e t t I R UC UL UAN UMB UA N B M

AB hnh d A I MRUN CU

LUABUr U B

1U U 2UCch v U =1U +2U41 b) ( )22Z= R Z ZL C+ =100 2O UAB = I.Z = 50 2 V c) Gin vect :1 gc lch pha ca uAN so vi i tanZL11 1R 4t= = =uMB tr pha so vi i 1 gc t/2 (v ZC >ZL). Vy uAN sm pha hn uMB 1 gc:312 4 2 4t t t t+ = + = . Da vo gin ta c:NB = 2MN v 14t=nn tam gic ANB l tam gic vung cn ti A v vy AB = AN Bi 2 Bi tp 2tr 174 SGK: Mt on mch in AB gm mt in tr thun R = 100O, mt cun cm thun v mt t in mc ni tip (Hnh 33.2). Bit in p hiu dng gia hai u in tr, cun cm, t in ln lt l UR = 50 V,UL = 50 V, UC = 87,5 V, tn s dng in l 50 Hz. a) Tnh t cm ca cun cm v in dung ca t in. B. Tnh tng tr ca on mch AB v in p hiu dng UAB C. V gin Fre-nen. Cn c vo gin : tm lch pha ca in p gia hai im A v N so vi in p gia hai im M v B; tm li UAB. Bi gii a) V I = R LU UR L=nn: Cm khng: ZL = LRURU = 100 O; t cm: L = LZ~e 0,318 H Tng t, ta c:- Dung khng: ZC = CRURU= 175 ;in dung ca t in: C = C1Z~e18,2 F. B. V AB l on mch RLC mc ni tip, nn c tng tr: Z = 2 2L CR (Z Z ) + = 125 OUAB = IZ = 50100.125 = 62,5 V C. Gin Fre-nen v nh Hnh 33.3, trong : AN R LU U U = + MB L CU U U = + Xt tam gic vung OSP c OP = UR, SP = UL, ta c: tanLRU1U = = , 14t=Gc to bi hai vect MBU , ANUl 132 4t t+ = .Vect ANUlp vi vect MBUmt gc 34t theo chiudng. Vy in p gia A v N sm pha 34t so vi in p gia M v B. 2 2ABU (OP) (OQ) = +

2 2U U (U U )R L AB C= +

2 2U = 50 +(50- 87, 5) = 62, 5VAB Bi 3:Cho mch in xoay chiu gm R,L,C ni tip .Cc in p hai u on mch : U = 120V ;2 u cun dy Ud = 120V ; hai u t in UC = 120V. Xc nh h s cng sut ca mch ? Gii : Ta v gin vec tc :U = 120V ;2 u cun dy Ud = 120V ; hai u t in UC = 120V. Thy tam gic OPQ u nn / 6 = =dPOH t ; / 6 = HOQ t / 3 t = 3cos2OHOQ = =Bi 4( TNPT 2008)Cho mch in RLC ni tip .Dng vn k nhit o in phiu dng 2 u on mch , 2 u t in v 2 u cun dy th s ch vn k tong ng l U ; UC ;UL .Bit U = UC = 2UL .Tnh h s cng sut ca mch? HNG DN ::tng t nh bi trn s :3cos2URU = =U2 = U2R +(UL-UC.2 = U2R +U2L= U2R + U2/4; Suy ra :32= U URC=45.10 ( ) Ft Bi 5: Mt cun dy dn c h s t cm L c mc ni tipvi mt t c in dung C ri mc vo 3 im A, B camt mch in xoay chiu c tn s f. o in p gia hai u on mch AB, gia hai u cun dy v C B L,r. A dU P O HI UCU Q 42 gia hai cc ca t in bng vn k c in tr rt ln,ta ln lt c: UAB = 37,5 V, Ud=50V, UC =17,5 V. o cng dng in bng mt ampe k c in tr khng ng k, ta thy I=0,1 A.Khi tn s f thayi n gi tr fm=330 Hz th cng dng in trong mch t gi tr cc i.Tnh t cm L, in dung C, v tn s f ca in p s dng trn. Hng dn gii : Gi s cun dy thun cm khng c in tr r th: UAB = UL UC = 50 17,5 = 32,5 V. Khng ph hp vi gi tr cho.Nn cun dy phi c in tr trong r ng k. Tng tr cun dy:2 2Z r Zd L= +Bin v gi tr hiu dng ca cng dng din c tnh theo cc cng thc: 0 002 2U UIZr Zd ddL= =+v2 2U UIZr Zd ddL= =+ Cng sut tiu th ca cun dy: P = Ud.I.cosd = I.r2 Vi h s cng sut: cosd= 2 2r rcosZZ rddL = =+hoccosrddUU = Ta tnh c:Tng tr ca cun dy: U 50Z 500I 0,1dd = = = O Dung khng ca t in: U 17, 5Z 175I 0,1CC = = = O .Tng tr ca on mch: U 37, 5Z 375I 0,1ABAB = = = O Khi f = fm, trong mch c cng hng (Imax) nn: em2 = 1LC 2 2 21 1 1LC=(2 f ) (2. .330)m me t t = =(1) Mt khc: ZAB2 = r2 + (ZL ZC.2 = r2 + ZL2 2ZLZC + ZC2 ZAB2 = Zd2 + ZC2 2ZLZC 2ZLZC = Zd2 + ZC2 ZAB2 = 5002 + 1752 - 3752 = 14.104

2.L. e .1C.e=4 4 4L L2 14.10 7.10 L=7.10 .CC C= = (2) Th (2) vo (1) ta c: 7.104.C2 = 21(2. .330) t

Suy ra: C=1,82.10-6 F; L=7.104.C=7.104. 1,82.10-6=0,128 H M: ZC = 1C.e=61 1 1f= 500C.2. f C.2. .Z 1,82.10 .2.3,14.175ct t = = Hz Bi 6: Cho mt on mch R, L, C mc ni tip, trong 410( ) C Ft= v L cun dy thun cm c t cm c th thay i c. t vo hai u mt on mch trn in pl200 U = (V), 50 (Hz) th thy khi ( )2L Ht= , in phai u cun cm l cc i. Hi: a.in tr ca on mch.b.in pcc i hai u cun cm. c.Cng dng in cc i khi L thay i. Phng php gin vct -p dng nh l hm s sin cho tam gicOAB ta c sin( ) sin( )OA ABOBA AOB=;sinos( ) sin cosLLU UU Uc OBA|| o= =Trong U v khng i ( )axULM khi sin l ln nht=> 2t| =-Mt khc: 2 2osCRcR Zo =+;Thay vo on biu thc trn ta c: I Ur Ud UL d43 C R B L A M N 2 2CL MaxR +Z(U ) = UR;Gi tr trn t c khi 2 2CLCR ZZZ+=Kt lun: Khi in phai u cun cm t gi tr cc i th: Gi tr cc i ln hn in phai u on mch v ln hn in phai u in tr v in dung. in phai u on mch nhanh pha hn cng dng in trong mach v lun vung pha vi in phai u in tr v cun cm. Gi tr cc i ca in phai u cun dy thun cm c xc nh bi: 2 2ax( )CL MR ZU UR+= t cm ca cun dy khi in phai u cun cm t gi tr cc i l: 2 2CLCR ZZZ+=Bi 7 :Cho mch in nh hnh v: UAB=120V;f =50Hz; R=30O ;L = H104t; cc vn k l tng (in tr khng nh hng n cc gi tr cn o). Tnh s ch ln nhtcu cc vn k V1 ;V2 khi C thay i ?:Hng dn :n 2C L2dd d) Z Z ( RZ . UIZ U += =,t khng i nn Ud ln nht khi mu nh nht khi :ZL -ZC =0=> V 200RZ . UUd(max) d= =; dng gin vect hay ohmca UC theo ZC => sin sin= =CU UkhongdoiM MAB;120sin0, 6 =CU MAB;V 200 U(max) C=Bi 8 : Cho mch in gm RLC khng phn nhnh ; Bit R =50 (O);); H (23Lt=t in c in dung C thay i , t vo gia 2 u on mch h..th xoay chiu ) V ( t 100 sin 2 240 u t =Ln lt iu chnh UR ; UL ;UC t gi tr cc i ? Tm cc gi tr cc i ? HNG DN: bng gin vectta c UR ;UL t cc i khi cng hng .Lc :URmax =240V ULmax =415V ,cn tm UCmax =>UCmax=480V . Bi 9(DH 2009):Mt on mch in xoay chiu gm in tr thun, cun cm thun v t in mc ni tip. Bit cm khng gp i dung khng. Dng vn k xoay chiu (in tr rt ln) o in p gia hai u t in v in p gia hai u in tr th s ch ca vn k l nh nhau. lch pha ca in p gia hai u on mch so vi cng dng in trong on mch lA. 4t. B. 6t.C. 3t.D. 3t . HNG DN ::dng gin vecto suy ra 4t =Bi 10: Cho mch in xoay chiu nh hnh v. Cun dy thun cm. Cho bit in p hiu dng gia hai im A, B l( ) V UAB200 = , gia hai im A, M l ( ) V UAM2 200 =v gia M, B l( ) V UMB200 = . Tnh in p hiu dng gia hai u in tr v hai u t in.Gii: Cch 1: Phng php vct buc (xem hnh a). + V( ) V U UMB AB200 = =nn tam gic MB ABU OUl tam gic cn ti O. Ch ( )22 22 200 200 200 = +nn tam gic l tam gic vung cn ti O. + Do tam gic MB RU OUcng l tam gic vung cn ti RU :2 1002= = = MBC RUU U . Cch 2: Phng php vct trt (xem hnh b). LUAMU |RU200 2 ABUo200 CU200 LU ORUI200 2200 A I 200 2CUMBUHnh ahnh b M dU Ud A B UC I M O LU N CU RU I |ABU o UL=2UC U U L C +ABU RU UC I44 + D thy( )22 22 200 200 200 = +nn AABM vung cn ti B, suy ra 045 = o = 045 |AMNB vung cn ti N2 1002= = = MBU UC R.S:2 100 = =C RU UBi 11: Cho mch in nh hnh v bn. in tr( ) O = 80 R , ccvn k c in tr rt ln. t vo hai u on mch mt hiuin th ( ) 240 2 100ABu cos t V t =th dng in chy trong mch c gi tr hiu dng) ( 3 A I = . in ptc thi hai u cc vn k lch pha nhau 2t, cn s ch ca vn k 2V l) ( 3 802V UV= . Xc nh L, C, r v s ch ca vn k 1V .Gii Cch 1:Phng php i sV2 ABAB MB AN MBU UZ ; Z v tan . tan 1I I= = = ( ) ( )2 2 240803L Cr Z Z + + = v ( )2280 33L Cr Z Z + = v. 180C L CZ Z Zr = ( ) 40 r = O ;( )2003LZ = O ( )803CZ = O( ) ( ) ( )32 3.10r 40 , L H , C F8 3= O = =t t;+ S ch caV1: ( ) V Z R I Z I UC AN V160 .2 21= + = = . Cch 2: Phng php vc t buc (xem hnh a). S dng nh l hm s cosin cho tam gic thng:( ) ( )23=3 80 240 23 80 3 80 + 240= 2 22. .cos0 030 = o 30 = + ( ) ( ) O = = = o =38080IUZ V tg U UCC R C.( ) ( ) O3200= = 200 = 2 3 80 + =IUZ V U ULL C Lsin + S ch ca Vn k V1: ( ) VcosUU URAN V1601=o= = . Cch 3: Phng php vc t trt. V gin vc t (xem hnh B.. Gi cc gc nh trn hnh. Theo bi ra:( ) V R I UR3 80 . = = . S dng nh l hm s cosin cho tam gic thng AABN: 233 80 . 240 . 2240. . 2cos2 2 2 2= = +=AM ABMB AM ABL AM NB C R V2 V1 45 0 0 060 90 30 = = = .M B A | 0 030 60 = = o+ Xt AAMN: ( ) ( )0C V1 0AMU MN AMtg30 80 V v U AN 160 Vcos30= = = = = = . + Xt AABG:( ) V sin . AB U GB U UC C L200 = + = + = . ( ) ( )200 21003 3LLUZ L L HItt= = O= = ;( ) ( )380 1 3.10100 8 3CCUZ C FI C t t= = O= =( ) O === = 40cos .IAM ABIAM AGIUrr. S:( ) ( ) F C H Lt 810 3=t 32=3 ., ,( ) O = 40 r , s ch vn k V1 l( ) V 80 . Bi 12: Cho mch in nh hnh v bn.Gi tr ca cc phn t trong mch ( ) ( ) r R F C H L 2 ,50,1= = =t t. in pgia hai u on mch( )0s100 u U co t V t = . in phiu dng gia hai im A, N l( ) V UAN200 =v in ptc thi gia hai im MN lch pha so vi in ptc thi gia hai im AB l 2t. Xc nh cc gi trr R U , ,0. Vit biu thc dng in trong mch. Gii: Phng php vc t buc (xem hnh c). + Tng t nh cch 2, ta thy tam gic OFE l tam gic u v G va l trng tm va l trc tm, suy ra: ( )030 , 200 = = = = V U U UAN C AB. + Tnh c:( ) V U UAB2 200 20= = . + Cng hiu dng:) ( 1200200AZUICC= = =+) (320030 cos 200 .32cos32320V U OH UAB R= = = = ) (3100), (3200O = O = = rIURR. T gin nhn thy, ABism pha hn ABul 6t.Vy, biu thc dng in:( ) A t i|.|

\|+ =6100 sin 2tt . S:( ) ) (3100), (3200, 2 2000O = O = = r R V U , ( ) A t i|.|

\|+ =6100 sin 2ttVN DNG : Cu 1 (Cu 56 24 cc kho th)Mch in xoay chiu gm t in v cun dy mc ni tip c mc vo in p xoay chiu c gi tr hiu dng U=150V.Dng vn k v ampe k nhit o ,ta c UC=70V; Ud=200V; I=2A . in tr cun dy bng bao nhiu ? A.10 O B.20 O C.40 O D.60 O HNG DN V gin vecto . Nhn tam gic OAB v tnhcos OAB sinOAB sin =rdUOABUUrr= Ur/I=60 O Cu 2(C2007)t vo hai u on mch RLC khng phn nhnh mt in pxoay chiu c u = U0cost. K hiu UR , UL , UC tng ng l in phiu dng hai u in tr thun R,cun dy thun cm L v t in C. Nu UR = .UL = UC th dng in qua on mch 46 R L C X ANB A. tr pha 4tso vi in p hai u on mch. B. sm pha 2t so vi in p hai u on mch. C. tr pha 2t so vi in p hai u on mch.D. sm pha 4t so vi in p hai u on mch. Cu3:tinpxoaychiuvohaiuonmchRLCnitipthccinphiudngcquanh 3 UR=3UL=1,5UC. Trong mch cA. dng in sm pha 6t hn in p hai u mch. B. dng in tr pha 6t hn in p hai u mch. C. dng in tr pha 3t hn in p hai u mch. D. dng in sm pha 3t hn in p hai u mch. Cu4:Chomchinxoaychiunhhnhv;cundythuncm.inphiu dng gia A v B l 200V,UL =38UR = 2UC. in p hiu dng gia hai u in tr R l: A. 180V.B. 120V .C. 145V.D. 100V. Cu 5 Cho mch in nh hnh v 1. Khi t vo hai u mch mt in pn nh c gi tr hiu hiu dng l 100V v tn s 50Hz v pha ban ubng khng th in p hiu dng gia hai u on mchAM l 60V v in p gia hai u on MB c biu thcuMB = 80 2 cos(100t + 4t)V. Biu thc ca in p gia hai u on AM l: A.uAM = 60cos(100t + 2t)V.B. uAM = 60 2 cos(100t - 2t)V. C.uAM = 60cos(100t + 4t)V.D. uAM = 60 2 cos(100t - 4t)V. Cu 6. Mch in xoay chiu gm in tr thun R=30( O) mc ni tip vi cun dy. t vo hai u mch mt in p xoay chiu u= 2 cos(100 ) U t t(V). in p hiu dng hai u cun dy l Ud = 60V. Dng in trong mch lch pha 6t so vi u v lch pha 3t so vi ud. in p hiu dng hai u mch (U) c gi tr A. 60 3(V)B. 120 (V)C. 90 (V) D. 60 2(V) DNG: BI TON V HP KIN (t gp ) A. Phng php: 1. Mch in n gin:a. Nu NBUcng pha viisuy raX ch cha 0Rb. Nu NBUsm pha viigc 2t suy ra X ch cha 0Lc. Nu NBUtr pha viigc 2t suy raX ch cha 0C2. Mch in phc tp: a. Mch 1 Nu ABUcng pha viisuy raXch cha 0LNu ANUv NBUto vi nhau gc 2t suy ra Xcha (0 0,L R ) b. Mch 2 Nu ABUcng pha viisuy ra X ch cha 0CNu ANUv NBUto vi nhau gc 2t suy raXcha (0 0,C R ) B. p dng1. Bi ton trong mch in c cha mt hp kn. V d 1: Cho mch in nh hnh v: L R A C Hnh 2 R C X ANB R L X ANB R L X ANB R L X ANB CLA BR MHnh 1 47 Bit :uAB = 200cos100tt(V) ;ZC = 100O ; ZL = 200O .I = 2 ) A ( 2; cos = 1; X l on mch gm hai trong ba phn t (R0, L0 (thun), C0) mc ni tip. Hi X cha nhng linh kin g ? Xc nh gi tr ca cc linh kin . Gii Cch 1: Dng phng php gin vc t trt. Hng dn B1: V gin vc t cho on mch bit + Chn trc cng dng in lm trc gc, A l im gc. + Biu din cc in puAB; uAM; uMN bng cc vc t tng ng. AB goc tai ACung pha iU 100 2( )ABUv+++ = Goc tai ATrepha 2200 2( )AMAMU soiU vt+++ = MN Goc tai M2U 400 2( )MNU Som pha soivt+++ = B2: Cn c vo d kin ca bi ton NBUxin gc v tr pha so vi i nn X phi cha Ro v Co B3: Da vo gin URo v UCo t tnh Ro; Co Li gii -Theo bi ra cos = 1 uAB v i cng pha.-UAM = UC = 200 2(V)UMN = UL = 400 2(V)UAB= 100 2(V) * Gin vc t trt V UAB cng pha so vi i nn trn NB (hp X) phi cha in tr Ro v t in Co. + URo = UAB IRo = 100 2 Ro = 100 250( )2 2= O+ UCo = UL - UC I . ZCo = 200 2 ZCo = 200 2100( )2 2= O Co = 41 10( )100 .100Ft t=. V d 2: Cho mch in nh hnh v uAB = 120(V); ZC = 10 3( ) O R = 10(O);uAN = 60 6 sin100 ( ) t v t;UAB = 60(v) a. Vit biu thc uAB(t) b. Xc nh X. Bit X l on mch gm hai trong ba phn t (Ro, Lo (thun), Co) mc ni tip Gii: a. V gin vc t cho on mch bit A Phn cn li cha bit hp kn cha g v vy ta gi s n l mt vc t bt k tin theo chiu dng in sao cho: NB = 60V, AB = 120V, AN = 60 V 3+XtthamgicANB,tanhnthyAB2=AN2+NB2,vyl tam gic vung ti N tgo = 60 160 3 3NBAN= =ACB N MXRUABUCURAMNBiUANUNBUR0Ul 0DUC0UR0UMNUAMNABUABMi48 6t= o UAB sm pha so vi UAN 1 gc 6t Biu thc uAB(t): uAB= 120 2 sin 1006ttt +| | |\ . (V) b. Xc nh X T gin ta nhn thyNBcho ln m trong X ch cha 2 trong 3 phn t nn X phi cha Ro v Lo. Do ta v thm c;0 0U UR L nh hnh v. + Xt tam gic vung AMN: 16 3URRtgU ZC Ct| | = = = =+ Xt tam gic vung NDB : 3cos 60. 30 3( )21sin 60. 30( )2U U VR NBOU U VL NBO||= = == = = Mt khc: UR = UANcos| = 6013. 30 3( )2v = ; 30 33 3( )10I A = =030 310( )3 3URORI= = = O 030 10 10 0,1( ) ( )3 3 3 100 3 3ULOZ L HL OI t t= = = O = =Vd3:Chomchinnhhnhv:BituAB=const;uAN=180 2 sin 100 ( )2t Vtt | | |\ ..ZC=90(O);R= 90(O); uAB =60 2 100 ( ) cos t V ta. Vit biu thc uAB(t) b. Xc nh X. Bit X l on mch gm hai trong ba phn t (RO, Lo (thun), CO) mc ni tip. Phn tch bi ton: Trong v d 3 ny ta cha bit cng dng in cng nh lch pha ca cc in pso vi cng dng in nn gii theo phng php i ssgpnhiukh khn.V d 3 ny cngkhc vd 2 ch chabittrcUABcnghaltnhchtcbittrongvd2 khngsdngc.Tuy nhintalibit lchphagia uAN v uNB, c th ni y l mu cht gii ton. Gii a. V gin vc t cho on mch bit AN.Phn cn li cha bit hp kn cha g, v vy ta gi sn l mt vc t bt k tin theo chiu dng in sao cho uNB sm pha 2t so vi uAN + Xt tam gic vung ANB * tano = 60 1180 3UNBNBAN UAN= = = o ~ 800 = 0,1t(rad) uAB sm pha so vi uAN mt gc 0,1t * 2 2 2U U UNB AB AN= + = 1802 + 602 ~ 1900 UAb = 190(V) ACB N MXRUABUCURAMNBiUANUNBUR0Uc0D49 biu thc uAB(t): uAB = 190 2 cos 100 0,12ttt t +| | |\ .= ( ) 190 2 cos 100 0, 4 ( ) t V t t b. T gin ta nhn thy NB cho ln m trong X ch cha hai trong 3 phn t trn X phi cha RO v LO. Do ta v thm c; U UR LO O nh hnh v. + Xt tam gic vung AMN: 90tan 190= = = = |URRU ZC C | = 450 UC = UAN.cos| = 180.2 90 290 2 2( )2 90UCI AZC= = = =+ Xt tam gic vung NDB 2 30 2cos 60. 30 2( ) 30( )02 2U U V RR NBO| = = = = = O| = 450 ULo = URo= 30 2 (V) ZLo = 30(O) 30 0, 3( )100L HOt t = =2. Bi ton trong mch in c cha hai hp kn Vd1:Mtmchinxoaychiucsnhhnh v.TronghpXvYch cmtlinhkinhocintr,hoccun cm, hoc l t in. Ampe k nhit (a) ch 1A; UAM = UMB = 10VUAB = 10 V 3 . Cng sut tiu th ca on mch AB l P = 5 6 W. Hy xc nh linh kin trong X v Y v ln ca cc i lng c trng cho cc linh kin . Cho bit tn s dng in xoay chiu l f = 50Hz. * Phn tch bi ton: Trong bi ton ny ta c th bit c gc lch (Bit U, I, P ) nhng on mch ch cha hai hp kn. Do nu ta gii theo phng php i s th phi xt rt nhiutr. hp, mt tr.hp phi gii vi s lng rt nhiu cc phng trnh, ni chung l vic gii gp kh khn. Nhng nu gii theo phng php gin vc t trt s trnh c nhng kh khn . Bi ton ny mt ln na li s dng tnh cht c bit ca tam gic l: U = UMB; UAB = 10 3 3 V UAM= tam gic AMB l A cn c 1 gc bng 300. Gii: H s cng sut:cosPUI =5 6 2cos2 4 1.10 3t = = = Tr.hp 1: uAB sm pha 4t so vi i gin vc tV: 3U UMB AMU UAB AM== AAMB l A cn v UAB = 2UAMcoso coso = 10 32 2.10UABUAM= coso = 30302o =a. uAB sm pha hn uAM mt gc 300 UAM sm pha hn so vi i 1 gc X = 450 - 300 = 150 X phi l 1 cun cm c tng tr ZX gm in tr thun RX v t cm LX A B MYaXiMURXULXKUABUYURYULYAHB450300150U50 Ta c: 1010( )1UAMZXI= = = O .Xt tam gic AHM: + 0 0cos1